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_ MATHEMATICS LIBRARY. 

















AN 


ELEMENTARY ALGEBRA 


FOR 


Setat and Academie’, 


BY 


JOSEPH W. WILSON, AM, 


PROFESSOR OF GEOMETRY IN THE CENTRAL HIGH SCHOOL OF PHILADELPHIA. 





PHILADELPHIA: 
ELDREDGE & BROTHER, 


No. 17 North Seventh Street. 





Entered, according to Act of Congress, in the year 1871, by 
ELDREDGE & BROTHER, 


4 
in the Office of the Librarian of Congress at Washington. 


Mie 


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SF ELECTROTYPERS, PHILAD’A. 


FERGUSON BROS. & CO., 
PRINTERS, PHILADELPHIA. 





PREFACE. 


2038300 


TINHE attention of Teachers is requested to the follow- 
ing points, in which the Author has endeavored to 
make this book a useful one: 


Clearness. The great aim throughout has been to make 
everything as plain as the nature of the subject would 
permit. Every principle has been explained, and unne- 
cessary verbiage avoided. 


A very Gradual Advance. Complex subjects have been 
subdivided as far as possible into their elements, so that 
the learner may have to take but one step at a time. 


A Practical Character. There is a continual review and 
repetition of whatever has been learned. Much more 
attention than usual has been given to Fractions, as a 
thorough drill on them is believed to be essential to a 
full comprehension of Algebraic operations, and a ready 
facility fn performing them. ‘The examples are very 
numerous, and are all original, so that this book may 
be used in connection with others, without fear of any 
other than accidental repetition. 

An Analytical Method. ‘This is essentially the same 


111 


iv PREFACE. 


as that which has been so successful in Mental Arith- 
metic. Synthetic arrangement has been subordinated 
to convenience of teaching. 


Accuracy of Language. Such expressions as Add the 
equations together, where the meaning is Add their corre- 
sponding members, have been carefully avoided. Hun- 
dreds of these errors have crept into some of the best 
books. The learner thus insensibly acquires a careless 
and incorrect use of language, which often clouds his 
apprehension of principles otherwise clear. 


In conclusion, an effort has been made to specially 
adapt the work to those teachers who are not satisfied 
with anything less than an ability on the part of their 
pupils to fully explain every operation which they per- 
form. To such the book is commended, in the hope that 
it will satisfy a need which the Author has himself fre- 
quently felt. 










@ 


S ENF 


oi 


SECTION PAGE 

I. THE FUNDAMENTAL SIGNS . d : ; : BS etd 

II. [ILLUSTRATIONS OF THE NATURE OF ALGEBRA . , hg AE 
III. PROBLEMS TO BE SOLVED BY EQUATIONS CONTAINING 

BUT ONE UNKNOWN QUANTITY . : ; ; rig 

IV. DEFINITIONS . : ‘ ; ; , ’ 3 f wea 

Penton? ne Re eI! : Gh Sot eee 


VI. PROBLEMS TO BE SOLVED BY EQUATIONS CONTAINING 


BUT ONE UNKNOWN QUANTITY . : : : . 28 

VII. SUBTRACTION q , : : : : F - . 30 

VIII. TRANSPOSITION . : ‘ : : : ; Pid d. 06 
IX. QUESTIONS PRODUCING EQUATIONS THAT REQUIRE 

TRANSPOSITION. : ; ; : : ‘ - 40 

X. MULTIPLICATION. THE SIGNS + AND — : : » 45 

XI. MULTIPLICATION OF LITERAL QUANTITIES : pb ~ ae 

XII. MULTIPLICATION. EXPONENTS : ; : ; . 48 


XIII. MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL . 61 
XIV. MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL 653 
XV. EQUATIONS REQUIRING MULTIPLICATION . : : . 58 


XVI. QUESTIONS PRODUCING EQUATIONS WITH ONE UNKNOWN 


QUANTITY ‘ : : : ‘ . ; ; emt 9 

XVII. Division. THE SIGNS + AND — : 4 : : a Oe 
XVIII. Division. LETTERS WITHOUTINDICES . ‘ A « 69 
XIX. DIVISION. MONOMIALS WITH INDICES . : : Od (4, 


1* y 


vi 


SECTION 


XX. 
XXII. 
XXII. 
XXII. 
XXIV. 


2. a 
XXVI. 
XXVIT. 


XXVIII. 
XXIX. 


XXX. 
2. @.5 
XXXII. 


XXXITI. 


De phe Now 


XXXV. 
XXXVI. 
XXXVII. 
XXXVIII. 
XXXIX. 


XL. 
XLI. 
XLIT. 
XLII. 
XLIV. 


CONTENTS. 


PaGE 
DIVISION OF A POLYNOMIAL BY A MONOMIAL . 5 aH 
REVIEW OF SUBTRACTION . ‘ : : ‘ . 78 
DIVISION OF A POLYNOMIAL BY A POLYNOMIAL x 7 
EQUATIONS WITH TWO UNKNOWN QUANTITIES. oe 
QUESTIONS PRODUCING EQUATIONS WITH Two UN- 
KNOWN QUANTITIES ~. . , . 94 
FACTORING. FUNDAMENTAL THEOREMS 96 
FACTORING. PRIME AND MONOMIAL FACTORS. . 100 
FACTORING. APPLICATIONS OF THE FUNDAMENTAL 
THEOREMS : : : ; : : . . 103 
FACTORING, (CONTINUED.) . ; ; : P . 106 
EQUATIONS WITH TWO UNKNOWN QUANTITIES, (CON- 
TINUED.) . 108 
THE LEAST COMMON MULTIPLE - 113 


REDUCTION OF FRACTIONS TO THEIR LOWEST TERMS 


REDUCTION OF WHOLE AND MIXED QUANTITIES TO 
FRACTIONS : ; ; F ‘ i : * 


REDUCTION OF FRACTIONS TO WHOLE AND MIXED 
QUANTITIES . ‘ é : : 2 " - 


REDUCTION OF FRACTIONS TO OTHERS HAVING THE 
LEAST COMMON DENOMINATOR . ‘ : ; 


ADDITION OF FRACTIONS ‘ ; ; , F - 
SUBTRACTION OF FRACTIONS 

MULTIPLICATION OF FRACTIONS . 
DIVISION OF FRACTIONS ‘ ‘ : x P F 


REDUCTION OF COMPLEX FRACTIONS TO SIMPLE 
ONES : : a : . : : 


CLEARING AN EQUATION OF FRACTIONS . . E 
QUESTIONS PRODUCING FRACTIONAL EQUATIONS 
REVIEW 

ELIMINATION * : : : F ; ; 


QUESTIONS PRODUCING EQUATIONS WITH Two UN- 
KNOWN QUANTITIES ; 


116 


120 


125 


128 
131 


« 132 
. 134 


136 


. 138 


141 


. 144 
. 151 


154 


ae 


SECTION 


XLV. 


XLVI. 


XLVII. 


XLVIII. 


CONTENTS. 


Vil 


Pace 


EQUATIONS CONTAINING MORE THAN Two UNKNOWN 
QUANTITIES é : : ; ‘ 4 ° : 


QUESTIONS PRODUCING EQUATIONS WITH MORE THAN 
Two UNKNOWN QUANTITIES . : - ¥ : 


PECULIAR EQUATIONS WITH MORE THAN Two Un- 
KNOWN QUANTITIES . 


PROBLEMS RELATING TO WORK 


XLIX. RATIO AND PROPORTION , . z : ; 7 : 


LXI. 
LXII. 
LXIII. 
LXIY. 
LXV. 
LXVI. 
LXVII. 


LXVIII. 
LXIX. 
LXX. 
LXXI. 


. ARITHMETICAL PROGRESSION 


. GEOMETRICAL PROGRESSION 


. ARITHMETICAL PROGRESSION, (CONTINUED.) . 


. GEOMETRICAL PROGRESSION, (CONTINUED.) . ; 
. QUESTIONS PRODUCING SIMPLE EQUATIONS . ‘ ‘ 
. INVOLUTION F ‘ 


. THE BINOMIAL THEOREM j S * 3 


EVOLUTION. Roots oF MONOMIALS 


SQUARE ROOTS OF POLYNOMIALS , 


. CLASSIFICATION OF EQUATIONS ; 5 ‘ d . 
. PURE QUADRATIC EQUATIONS AND OTHERS SIMILARLY 


SOLVED 
EQUATIONS WITH TWO UNKNOWN QUANTITIES 
ADFECTED QUADRATIC EQUATIONS 
MISCELLANEOUS EXAMPLES. REVIEW QUESTIONS ° 
NEGATIVE AND FRACTIONAL EXPONENTS 7 ; : 
RADICALS . : F . : P r ‘ - : 


REDUCTION OF RADICALS TO THEIR SIMPLEST FORMS . 


164 


168 


ay bp 
anne d 


183 


- 192 


ELIMINATION OF RADICALS FROM THE DENOMINATORS — 


OF FRACTIONS - . - A : 5 ; : 
REDUCTION OF RADICALS TO A COMMON INDEX . ‘ 
ADDITION AND SUBTRACTION OF RADICALS 
MULTIPLICATION AND DIVISION OF RADICALS 


AN ALGEBRAIC PUZZLE 


Now Ready. 
Saw Pete 


A KEY TO WILSON’S ALGEBRA, for the use 
of Teachers,» . . .*% =o. 8° R°e2 Prigegiaaa 


viii 





ELEMENTARY ALGEBRA. 


save Eatea— 


SECTION IL. 
The Fundamental Signs. 


THE sign of Additionis +. Itiscalled plus. Thus,4+6 
fs read four plus six; and it shows that four and six are to 


be added. 


The sign of Subtraction is—. It is called minus. Thus, 
7 — 2 is read seven minus two; and it shows that 2 is to be 
subtracted from 7. 


The sign of Multiplication is x. Thus, 5 x 4 means 5 
multiplied by 4. 


The sign of Division is--. Thus, 15-3 means 15 divided 
by 3. A fraction also indicates division. Thus, 1 means 
15 divided by 3. 


The sign of Equality is =. Thus 4+ 7 = 8 + 3 is read 
four plus seven is equal to eight plus three. The expression 
that two things are equal is an equation. Thus, 4 + 7 = 
8 + 3 isan equation, Sois16—2=—>3+14 10. 


Other signs will be explained when there is occasion to use them. 
9 


10 ELEMENTARY ALGEBRA. 


SECHGNeTI. 
Illustrations of the Nature of Algebra. 


Tuer Mathematics are the sciences which treat of number- 
ing and measuring. For example, Arithmetic explains how 
to calculate by means of numbers, and Geometry shows how 
to measure things, at first by placing them side by side, and 
afterward in many other ways. 

Algebra is that branch of mathematics in which numbers 
are represented by letters. 

The different ways in which numbers are represented by 
letters, and the use of doing so, cannot be fully explained 
until a great deal of progress is made in the study of algebra; 
but some preliminary ideas of the science may be obtained 
by considering the following questions: 


ProsiEM 1. I wish to divide 21 marbles between Henry 
and Charles, so as to give Henry twice as many as Charles. 
How many must I give to each? 


Soxtution. I wish to give to Charles a certain number, 
and to Henry twice that number. Hence, I must give to 
both ‘of them three times that number. Then, since both 
together are to have 21, I know that three times that number 
are 21. Therefore, once that number is 1 of 21, which is 7. 
Hence the number which Charles is to receive is 7; and 
twice that number is 14, which Henry is to receive. 

If I had used the letter n, instead of the words a certain 
number, the work would have been somewhat shortened. 
Thus: 


Srconp Soxution. Let the number that I wish to give 
to Charles be called n; and to Henry 2n. Then, since both 
together are to have 21, I know that 3n = 21. Therefore 
once n = 7, which is Charles’s share; and 2” = 14, which 
is Henry’s share. 

It will be observed that at the beginning of the solution 


ELEMENTARY ALGEBRA. 11 


the number of marbles to be given to Charles is not known. 
In the first solution it was called a certain number; in the 
second, it was called n. Now for more than two hundred years 
mathematicians have been in the habit of using the last 
letters of the alphabet, x, y, z, u, etc., to stand for numbers 
that are not known. If we adopt their practice, the solution 
just given may be changed to the following: 


Turrp Soturion. Let x represent the number of mar- 
bles for Charles. 


Then 2 represents the number for Henry. 


Hence, oe = 21, 
_ Dividing by 3, “== 7. 
Whence, Q2¢ = 14, 


Therefore Charles is to receive 7 marbles, and Henry 14. 


PrRoBLEM 2. What number added to itself will make 16? 


SoLution. Let x represent the number. 


Then, : e+a2= 16. 
Collecting, 2a = 16: 
Dividing by 2, % ==, Os 


Therefore 8 is the number which when added to itself will 
make 16. 


ProsiEeM 38. In a flock of 72 sheep, there are seven 


times as many white ones as black ones. How many are 
there of each kind? 


Sotution. Let # stand for the number of black sheep. 
Then 7 x stands for the number of white ones. 


Hence, G iT ae -72: 
Collecting, 8a = 72. 
Dividing by 8, ve oe 

Whence, Te = 63. 


Therefore there are 9 black sheep, and 63 white ones. 


12 ELEMENTARY ALGEBRA. 


-Prosiem 4, William had three times as much money 
as James, and the difference between William’s money and 
James’s was 20 cents. How much had each? 


Sotution. Let 2 stand for James’s money. 
Then 8 stands for William’s. 


Hence, 32—2 = 20. 
Collecting, 2% == ZU; 
Dividing by 2, z= 10, 
Whence, 32 = 80. 


Therefore James had 10 cents, and William 30. 


PROBLEM 5. It is required to divide $600 among three 
soldiers, so that one, who is a lieutenant, may have twice as 
much as another, who is a corporal, and the third, who is a 
captain, may have as much as the corporal and the lieuten- 
ant together. 


SotutTion. Let x represent the corporal’s share. 
Then 2 x represents the lieutenant’s share. 
And 32 represents the captain’s share. 


Hence, a+2a2+ 32 = 600. 
Collecting, 6 2 =760U, 
Dividing by 6, xz == 100. 
Whence, 22 = 200. 
And, 82 = 800. 


Therefore the corporal should receive $100; the lieutenant, 
$200; and the captain, $300. 


In the preceding solutions, one of the quantities to be 
found has always been represented by x. The problems are 
such that when one of the unknown quantities is found, the 
others can be very easily obtained by multiplication, or 
some other simple operation. Such problems can always be 
solved by using a single letter, and the equations obtained 


ELEMENTARY ALGEBRA. 13 


from them are called equations containing but one unknown 
quantity. 


There are some questions which cannot be easily solved without 
using two or more letters. They will be explained in a subsequent 
part of this book. 


In the fifth problem, there are three things to be found, 


and we might have represented any one of them by z. 
Thus: 


4 
SECOND SOLUTION OF PROBLEM 9. 
Let x represent the lieutenant’s share. 


Then 4 represents the corporal’s share. 
And 13 represents the captain’s share. 


Hence, zx+%tx2+ 142 = 600. 
Collecting, 8a = 600 
Dividing by 3, a = 200, 
Whence, + £2 = 100. 
And, 1} a2 = 300. 


Therefore the lieutenant should receive $200; the corpo- 
ral, $100; and the captain, $300. 


THIRD SOLUTION OF PROBLEM 5. 

Let «x represent the captain’s share. 

Then 42 represents the corporal’s share. 
And # represents the lieutenant’s share. 


Hence, xc+é¢a+ ea = 600. 
Collecting, 2a == GUE 
Dividing by 2, x = 800. 
Whence, +x = 100. 
And, $2 = 200. 


These solutions are not quite as easy as the first, because 
they contain fractions. In equations of this kind, it is gen- 
erally best to represent the smadlest of the unknown quan- 
tities by x. By doing so, fractions will be avoided. 

2 


14 ELEMENTARY ALGEBRA. 


SECTION III. 


Problems to be solved by Equations containing 
but One Unknown Quantity. 


1. JoHN and George have each the same number of 
apples, and both together have 36. How many has each? 
Ans. 18. 


2. If 67 = 30, what is x equal to? Ans. # = 5. 


3. Ann has a certain number of books, and Jane has four 
times as many. They together have 25 books. How many 
has each? Ans. Ann has 5, and Jane has 20. 


4. In a brigade of 2233 soldiers there were 10 times as 
many private soldiers as officers. How many private sol- 
diers were there? Ans. 2030 private soldiers. 


5. A and B together are worth $2432, and A is worth 7 
times as much as B. How much is B worth? Ans. $304. 

6. What number added to itself amounts to 88? Ans. 19. 

7. William and Robert caught 81 fish, and Robert caught 


8 times as many as William. How many did Robert catch? 
Ans. 72 fish. 


8. What number added to 5 times itself will amount to 72? 
Ans. 12. 

9. A farmer bought a horse and carriage for $420, and 
the carriage cost 3 times as much as the horse. What was 
the cost of each? Ans. The horse cost $105; and the car- 
riage, $315. 

10. At a county election, there were two candidates for 
sheriff. One received 3 times as many votes as the other, 
and his majority was 220. How many votes had each, and 


how many were polled in all? Ans. One received 330 
votes; and the other, 110; in all, 440. 


11. Ifa+32+ 52 = 63, whatisthe value ofx2? Ans. 7. 
12. Thomas travels a certain distance one day, and twice 


ELEMENTARY ALGEBRA. 15 


as far the next; in all 42 miles. How far does he travel 
each day? Ans. 14 miles the first day, and 28 miles the 
second. 


13. In a store-room containing 45 casks, there are 4 times 
as many full as empty. How many are full? Ans. 36. 


14. The angles of a triangle always amount to 180 de- 
grees. If one angle of a triangle is twice a second, and the 
third is 3 times the second, how many degrees are in each? 


Ans. 60°, 30°, and 90°. 


Nore. Let x = the number of degrees in the second angle: then 
2x = the number of degrees in the first, and 32 in the third. 


15. If 2 angles of a triangle are equal, and the remaining 
angle is twice their sum, how many degrees are in each? 
Ans. 30°, 30°, and 120°. 


16. If one angle of a triangle is 3 times a second, and the 
third is twice the first, how many degrees are in each? Ans. 
54°, 18°, and 108°. 

17. The angles of a quadrilateral, or four-sided figure, 
always amount to 360°. If they are equal, how many de- 
grees are in each? Ans. 90°. 

18. If one angle of a quadrilateral is twice a second, the 
third twice the first, and the fourth equal to the third and 
second together, how many degrees are in each? Ans. 60°, 
30°, 120°, and 150°. 

19. Two men start from the same point and travel in op- 
posite directions. The first travels 8 times as fast as the 
second. How far has each travelled when they are 48 miles 
apart? Ans. The first has travelled 36 miles; the second, 
12 miles. 

20. Mary is 14 years older than Jane, and her age is 8 
times Jane’s. How old is each? Ans. Mary is 16 years 
old, and Jane is 2 years old. 


Notre. Let += Jane’s age: then 8x— Mary’s. We have, by the 
conditions of the question, 8% —z = 14, 





16 ELEMENTARY ALGEBRA. 


SECTION IV. 


Definitions. 


A monomial is any algebraic expression which stands by 
itself, and is not composed of parts united by the sign + or 
—. Thus, z isa monomial; — 32 isa monomial; 14a is 
a monomial. 


A polynomial is any algebraic expression which is com- 
posed of parts united by the sign + or —. Thus, a + 6 is 
a polynomial; 2<— 3y is a polynomial; a —3b + 2n 
—e isa polynomial. 

The parts of a polynomial are called terms. A monomial 
consists of a single term. 

A polynomial of two terms is called a binomial. One of 
three terms is called a trinomial. Thus, 2a — 361s a bi- 
nomial; —8a2+a-+ 4n is a trinomial. 


A positive term is one having the sign + before it. A 
negative term is one having the sign — before it. In the 
polynomial x + 2a — 32 —ce-— aa, the first term is posi- 
tive, because the sign + is understood before it; the second 
term is also positive; the other three terms are negative. 


Sometimes a dot is used between letters as a sign of mul.- 
tiplication. Thus, a.6 is the same asa X 6. Multiplica- 
tion is also frequently denoted by writing two letters together. 
Thus, ab is also the same as a X 6. The same is the case 
with a figure and a letter. Thus, 3x is the same as 3 X &. 
The figure before the letter is called a coefficient. Thus, in 
the. term 7 n, the coefficient of n is 7. 

It must always be remembered that in algebra, letters 
stand for numbers. The expression ab means that the 
number which a represents is to be multiplied by the num- 
ber which } represents. Sometimes a stands for one num- 
ber, and sometimes for another. 


When a number is multiplied by itself oue or more times, 


ELEMENTARY ALGEBRA. 17 


the product is called a power. The second power of 5 is 25, 
because 5 X 5== 25. The third power of 5 is 125, because 
5x5xX5—125. The third power of « is xxx. The 
fourth power of ais aaaa. The second power is often called 
the square, and the third power the cube. 

Instead of writing aaaa, we generally write a‘. The 
number 4 is used to show how many times a is taken as a 
factor. Such a number, written to the right of the upper 
part of any letter, is called an index or exponent. The 
power 6bbb06 is generally written 6°, where 6 is the exponent. 
We do not often write 1 as an exponent. Thus ' is just 
the same as #. It is often said that the exponent of a letter 
written without any, is 1 understood. Thus, the exponent 
of x, or a, or c, is understood to be 1. 


A root is the quantity which is multiplied by itself one or 
more times to forma power. Thus 2 is the root of the 
powers x’, x*, x‘, etc. It is the second root of x’, the third 
root of x’, the fourth root of zx‘, etc. The second root is 
often called the square root, and the third root the cube root. 
The square root of 64 is 8, because 8 x 8= 64. The cube 
root of n? is n, because nnn = n’. 

The radical sign, /, shows that some root of the quantity 
placed under it is to be taken. To denote the square root, 
it is used alone. To denote any other root, a small figure 
is placed over it. Thus, ./9 means the square root of 9, 
which is 3; and ~/8 means the cube root of 8, which is 2. 


When several ‘terms are to be considered together, they 
are placed within a parenthesis, or a vinculum is placed 
over them. Thus, (5+ 2+ 7) x 9,or5+2+7 x 9, 
shows that the sum of 5, 2, and 7 is to be multiplied by 9. 
If the expression were 5+ 2+7 x 9, the 7 alone should 
be multiplied by 9 before adding it to the sum of 5 and 2. 

Operations of multiplication and division must be performed 
before those of addition and subtraction, unless parentheses in- 
dicate otherwise. 


ae B 


ELEMENTARY ALGEBRA, 


EXAMPLES. 


Find the value of each of the following expressions: 


1. (8+ 3) x 2. Ans. 22. 
2. 8 + 3.2, Ans. 14. 
3. (8 — 3) X 2. Ans. 10. 
4,.8—38 x 2. Ans. 2. 
5. (6 + 4) x 3. Ans. 27. 
6. (5 + 3) x 9. Ans. 72. 
7. (8 +6+3) x 2. Ans. 34. 
8. (8+3—5)x 4 Ans. 24. 
9. (10 + 8 — 2) x 7. Ans, 112. 
10. 10 + 8—2 x 7. Ans. 4. 
11. (8 —3 + 4) x 3. Ans. 27. 
12.8—3+4 x 3. Ans. 17. 
13. (6 —2 + 3) x 4. Ans. 28. 
14.6—2+3 x 4. Ans. 16. 
15. 8 x (5 — 2). Ans. 24. 
16. 8 x 5 — 2, Ans. 38. 
17..9 x (7.—*4). Ans. 27. 
18.9 x 7 —4. Ans. 59. 
19.5 x (8 + 38 — 7). Ans. 20. 
20.5x 8+ 3—7. Ans. 36. 
21. 7 x (9.+ 2). Ans. 77. 
22.7 xX 9+ 2. Ans. 65. 
23. 2X (7 —3-++ 8). Ans. 24. 
24.2x 7—8-+ 8. _ Ans. 19. 
25.9 x 4+ 2 -+ 8. Ans. 41. 
26.9 X (4+ 2 +4 8). Ans. 81. 
27. (4+ 3) x (8 — 5). Ans. 21. 
28. (4+ 3) x 8 — 5. Ans. 51. 
29.4+3 x (8 —5). Ans. 138. 
30.4+3 x 8—5. Ans, 23. 
31. (2+ 3) x 4. Ans. 20. 
-82. (2 + 3) x (4 + 8). Ans. 35. 


ELEMENTARY ALGEBRA. 19 





33. (6 — 5) x 4. Ans. 4. 
34. 6 x (5 — 4). Ans. 6. 
30.6 x 4— 9. Ans. 19. 
36. (6 + 4) + 5. Ans. 2. 
37.6 + 4 + 5. Ans. 64. 
38. (2 +3—4+ 5) x 6. Ans. 36. 


3 
39. (24+ 38 — 4) +54 5. Ans. 5 
40. (2+3—4)+(5+4 5). Ans. 3 
41.2+3—4—5+4 5. Ans. 9 
42.2+3+4— (5+ 4). Ans. 5 
43.2+3—4+4+5-x 6. Ans. 
44.2+3 x (—4-+ 5) = 5. Ans. 
45.5—2+11 x 3—2. Ans. 
46.5—2+11 x (8— 2). Ans. 14. 
47. (5— 2) x 114+ 3—2. Ans. 34. 
48.7+19—3 x 4—8383. Ans. 11. 
49.7 + (19—3) x 4—38._ Ans. 68. 
50. 7 + (19 — 8) x (4— 3). Ans. 23. 


C2 bo 


co 
ee Se en eee 


QUESTIONS ON THE FOREGOING DEFINITIONS. 


Why is —a + 2 —e a polynomial? What kind of a polynomial 
is it? What kind of a polynomial is a+ 5? 

Is ab a polynomial or a monomial? Why? Is 127 atr3y? a poly- 
nomial or a monomial? Why? 

How many terms are in the polynomial mn? — 7 abe? + x — 32 yz 
+ w8? Which terms are positive, and which are negative? Name 
all the coefficients, and all the indices. (Some are 1 understood.) 

Write ce + ¢+e¢+e¢-+cina shorter way. 

Write ececece in a shorter way. 

In the power 79, what is the index of ¥? 

In the monomial 17 acz*, what is the coefficient of acz4? What is 
the index of a? (Ans. 1 understood.) What is the index of c? of x? 
Is za power? What is its fourth root? 

In the monomial z°7?, what is the coefficient of z5y3? What is the 
exponent of x? of y? 


What is 4/z? equal to? What is ~/2zt equal to? What is 4/16? 


What is %/27? 
What is the fifth power of 2? of 1? 


20 ELEMENTARY ALGEBRA. 


SECTION V. 
Addition. 


In algebra, we may add terms together by writing them down 
one after another with their proper signs. Thus, if we wish 
to add x to y, we may write the sum either # + y ory + a. 
It does not matter which is placed first. The sign + is un- 
derstood before x or y when no sign is expressed. To 
add x to — y, we write either x — yor —y+ a. Addition 
in algebra includes more than addition in arithmetic, since 
we may find the sum of both positive and negative terms. 


Similar terms are those which do not differ otherwise than 
in their coefficients. Thus 3a is similar to 5a: the co- 
efficients, 3 and 5, are different; but the rest of the terms is 
the same. So, 46, 60, 6,46, and 120, are similar; but 3a 
and 5 6 are dissimilar, as the letters differ. Again, w’y and 
2x?y are similar; but wy and 2.y’ are dissimilar. Since 
either xy or yx means the product of x and y, they are not 
different, and are consequently similar. So 3ad is similar 
to 14 ba. 


Similar terms may be added into asingle term. Thus 
the sum of 8aand 4ais 7a. Yoadd similar terms, perform 
the operations of addition and subtraction on their coefficients. 


ProsiEm 1. Add 5 apples, 7 apples, and 16 apples. 
Prosiem 2. Add 5 dozen, 7 dozen, and 16 dozen. 
ProsieM 3. Add 5a, 7a, and 16a. 


OPERATIONS. 
1. 5 apples, 2. 5 dozen, 3. 5a, 
7 apples, 7 dozen, 74, 
16 apples, 16 dozen, 16a, 

















28 apples. 28 dozen. 28 a. 


ELEMENTARY ALGEBRA. 21 


ProsuiEM 4. John has 6 cents in his pocket, 9 cents at 
home in a drawer, and his brother owes him 7 cents. What 
is his financial condition ? 


OPERATION. EXPLANATION. The sign + is understood 
6 cents, before every-term in this operation and the 

9 cents, three preceding ones. The sign + is often 

7 cents, used before terms which express how muck 
——- a person has, or how much is due to him. 
22 cents. The result shows that John is worth 22 cents. 


ProsiEM 5. William has no money, and he owes a 
grocer 8 cents, a pieman 4 cents, his sister 3 cents, and a 
playmate 10 cents. What is his financial condition? 








_ OPERATION. 

— 8 cents, 

— 4 cents, EXPLANATION. The sign — is often 

— 3 cents, used before terms which express how 

— 10 cents, much a person owes. William owes 25 
cents, and has no money to pay it with. 

— 20 cents. 


PropiEeM 6, A boy has 16 cents, and owes 9 cents. 
What is his financial condition ? 


OPERATION. EXPLANATION. If he pay his debt, he 
16 cents, will have left the difference between 16 
— 9 cents, and 9cents. His true financial condition 


is the possession of 7 cents, clear. 








7 cents. 


ProsieM 7. A boy has 11 cents, and owes 17 cents. 
What is his financial condition ? 

OPERATION. EXPLANATION. He owes more than he 

11 cents, possesses. If he pay as much as he can 

-— Li cents, of his debt, it will be diminished by all 

he possesses, but he will remain 6 cents 

— 6 cents. in debt, without anything to pay it with. 








p pe ELEMENTARY ALGEBRA. 


Propiem 8. George has 63 cents in his money-box, 10 
cents in his pocket, and William owes him 10 cents. He 
owes his brother John 7 cents, and a confectioner 14 cents. 
What is his financial condition ? 





OPERATION. 

63 cents, 

10 cents, EXPLANATION. The positive terms 

10 cents, amount to 83 cents, and the negative 
— 7 cents, ones to 21 cents. Their difference is 62 
— 14 cents, cents, which George is really worth. 

62 cents. 


PRopLEM 9. James has 6 cents in his pocket, 18 cents 
at home, and another boy owes him 10 cents. He owes 12 
cents to a schoolmate, 14 cents to a stationer for a copy- 
book, and 11 cents’ to an apple-woman. What is his finan- 
cial condition? 


OPERATION. 
6 cents, EXPLANATION. The positive terms 
18 cents, amount to 34 cents, and the negative 
10 cents, ones to 87 cents. Their difference is 3 
— 12 cents, cents; and as the negative sum is the 
— 14 cents, greater, James owes 3 cents more than 


— 11 cents, he has money to pay his debts with. 








— o cents. 


PropuiEem 10. Add —a, —7a, —a,—38a, —7a, —8a, 
and — 2a. 


ProsiEeM 11. Add 3a, — 7a,4a, 6a, 9a, —2a,—11a, 
and 4a. 


-ProspiteM 12. Add 8a, —4a, —14a, 3a, — 5a, 6a, 
— 4a, and — 6a. 


ELEMENTARY ALGEBRA. 23 











OPERATIONS. 

10. — a, LP 3a, 12. 8a, 
—= 7 a, — 7a, — Aa, 
— 4a, 4a, — 14a, 
— 3a, 6 a, 34, 
— Ta, 9 a, — 5a, 
— 8a, — 2a, 6 a, 
— 2a, — lla, — Aa, 

4a, — 6a, 
— 29 a. 
6 a. — l6a. 


Thus we see that in addition it is convenient to write 
similar terms in a single column, find the sum of the positive 
terms, then of the negative terms, take the difference of these 
sums, and prefix the sign of the greater sum. The operation 
of counting is performed on the coefficients only. 

The addition of similar terms is often called collecting 
them. 


ProspuiEeM 13. Collect 6a —26 + 5a—36+a— DB. 


Prosiem 14, Collect a’ — a’y + ax? — 38a°x + ax’y 
— 4ax’* + a'r. 








OPERATIONS. 
13. 6a— 26, 14. axe—axry+ ax’, 
5a — 8b, — 8ar 4+ xy — 4a2’, 
a— Ob, an 
124 — 66. — we — 3 ax’, 


ProsptEM 15. Add 7Ta—3b, b— 2a, —38a-+ 20, 
76 —a, and 206. 


PrositeM 16. Add 38n—m + 2,2 +n, 3n —m— 2, 
m + 2, and — 4. 


24 


15. 


CONDO > 


ELEMENTARY ALGEBRA. 








OPERATIONS. 
ta 356; 16. 3n—m-4 2, 
—2a+ 8, n + 2, 
— 3a+ 28, 3n—m— 2, 
= ted 0; m + 2, 
vi oe 4, 
a+90b. Tn — mm. 
EXAMPLES. 
. Collect a, a, a, a, a, a, a, and a. Ans. 8a, 
. Collect — x, — x, — x, — x7, — x, — x, and — a. 


Avia a 


. Collect a (a + y), 2a(a@+ y), and 8a (a+ y). 


Ans. 6a(x% + y). 


. Collectt5a + 32+22-+ Tx + a. Ans. 182. 
. Collect —382—42—92x—38a4. #£xAns. —192. 
. Collect Ty — 2y—5y + 4y. Ans. 4y. 
. Collect Vy —4y + By + 15y—13y. Ans. 15y. 
. Collect 15z — 12z + 182 — 52z — 82 — 122. 


Ans. — 42, 

. Collect 9a —za2 + 10x — 82+ 38x—Aza. 
Ans. 9 x. 
. Collect —x + 2a. Ans. X. 
. Collect « — 2 2. Ans. — xX. 


. Collect 82 — lla 4+ 18a—152+4 10a. Ans. 52, 
. Collect 5a—Ta+ 8a—9a—12a+4a. 


Ans. — 19a. 

. Collecta—a+ 2a—2a+ 9a — l6a. 
Ans. — Ta. 

. Collect 36 —5b + 76—105—64 38. 
Ans. — 3b. 
. Collect — b —2b— 306. Ans. — 6). 
. Collect — b—26+4 36. Ans. 0. 


. Collect ax + Yax — Tax + 38axr — ax + Baz. 


Ans. 10 az. 


ELEMENTARY ALGEBRA. 25 


19. Collect xy —- 8 ay + Tay. Ans. 0. 
20. Collect (2 + a) + 19 (# + a) — 138 (% 4 a). 

Ans. 7 (« + a). 
21. Collect x —184a+ 19a—182+4+19%+4 a. 


Ans. Tx + Ta. 
Poecolec 2 —106 + Ta — ba — 26+ 5 — 8a. 
Angi:r— 10-0: 


23. Collect y + 11z— 13 y + 21z—15z + 4z—D5y. 
Ans. —17Ty + 212. 

24. Collect 5ba — 2 b’x + 3 b’x — 2 bz. 
Ans. 3ba + Bx. 
25. Collect 18 ay — 2arz 4+ 1342 — 5ay — 11 zz. 
Ans. 13 ay. 
26. Collect 18a? + 9a — 14a — 8a? + 6a + a. 
Ans. 6@ + a. 
27. What is the sum of 6 ab, 6 ab, — 6 ab’, and 5a’b?? 
Ans. 6a7b + 6 ab — 6 ab? + 5 a’b?. 
28. What is the sum of 7 a’2’y, — 8a’xy + 18 a’2*y, and 


— 8 ax’y? Ans. 17 a’2*?y — 8 ary. 
29. Add x —yand x + y. Ans.. 2x. 
30. Add y and — y. Ans. 0. 


31. Add, as, a’x, ax’, a?x’, and az. 
Ans. 2ax + aa + az? + a?x’. 
32. Add 2a to a’. Ans. 2a + a’. 
33. Add ab and ab. Ans. 2 ab. 
34. Add abe, — 3 abc, 4 abe, and — 3 aed. 
Ans. 2abe — 8 aed. 
35. Add a — 3y, 6a — Dy, and Ta — 3y. 
Ans. 14a —11y. 
386. Add 10” +4 38y,5a”—2y, and 84 —4y. 
| Ans. 18a — 38 y. 
37. Add 9 be — 7x + z, 8be — 3 a, and 5 be. 
Ans. 22 be —10 x + z. 
38. Add 10 ay + 2 — abr, — 8 ay + 32, Tay + 
5 abx, — 8ay —22,and 622—4abzr. Ans. Bay+ 82. 
39. Add 7x and 7 y. Ans. Tx + Ty. 
38 


26 ELEMENTARY ALGEBRA. 


40. Add 52+ a,—3a + 6, and—2a—a,. Ans. 6. 
41, Add 13 by + 6ac, —11 by + 5ac, and — 2 by — 
11 ae. | Ans. 0. 
42. Add 21 ax + 18 abe — abe + 2ay, 18 abe 4+ 14 ax 
— ax — day, and 3 ay — 27 abe + Sax. 
Ans. 87ax — 2 abe. 
43. Add 3ed — 2de + ed — ed’? + ed’, 5ed — e’'d + 
3 ed? — ed’, 7 cd — Tde + 8 cd’, 5e’'d — 2de + 3e’d’, and 
21ed + 8de,—5ed + 2cd’ — 11 ed’. 
Ans. 36 ed — &8de + 7 cd? — 8 ed’. 
44, Add 13 ab + 2ed and 25a’b + 8ed. 
Ans. 13ab + 25 ab + 5 cd. 


45. Add aba, bex, and bea. Ans. aba + 2 bex. 
46. Collect into one sum 8 bex, — 2 bex, 3bex, and — 
5 bex. Ans. — bex. 
47. Find the value of 3ryz — 2ayz + Sayz — 4axyz + 
LYZ — yz. Ans. 2 xyz. 


48. Add 18atryz + 21a — 38y, —3av + 2y + 
19 atxyz, — 3a’ + 2y — 2atxyz, and datryz — 3a — 
Qy + 12atryz. Ans. 52 atxyz + 12a + 21y. 

49. Find the value of 21 a’z*y, 3 atay? — 13 a®aty + 
2a°x°y, 13 atxcy*? — 11 aa’y + 8 a®a®y, 12 atay? — 10 a®ay, 
and 5 a?x°y. Ans. a’a'y + 28 atxy? — 4 a’ay. 

50. Add 9 a’b3ct + 8 a3b’ct — 7 atb?c*, 21 a®b’ct — 5a’bict, 
— ll a’b'ct + 2 a3b’ct, 21 a*b’c®, and 5 a®b’ct — 2 a®b’ct. 

Ans. — 7 a’b'ct + 34 a8b’ct + 14 a407e*. 

51. Add 9(a+ y), — 14 (a+ 2), and — 9 (a+/y). 

Ans. — 14 (a+z). 


52. Add ae and ca. Ans. 2 ae. 
53. What is the sum of 3a) — 5 ba + 12 ab — ba. 

Ans. 9 ab. 
54. Add abs, 2 aab, and — bxa. Ans. 2 abz. 


55. Add cay, —9 cyx, and —4ayc, Ans. —12 cay. 


Norse. In the last four examples, the terms are similar in each 
example. The letters are the same, only differently arranged. 


ELEMENTARY ALGEBRA. 27 


FURTHER ILLUSTRATIONS OF ADDITION. 


ProsLEeM 1. A mass of iron and wood is sunk in water. 
The iron weighs 16 Ibs. under water; and the wood would 
buoy up 13 lbs. What does the whole mass weigh while 
under water ? 

OPERATION. 
16 lbs. EXPLANATION. Consider weights posi- 

— 13 lbs. tive quantities. Then what buoys up or 
diminishes a weight must be a negative 
3 lbs. quantity. 





ProsieM 2. If the iron in the above problem had 
weighed 24 Ibs. under water, and the wood had been able to 
buoy up 30 lbs., what would the whole mass have weighed 
under water ? 


OPERATION. 
24 Ibs. EXPLANATION. The result here shows 
— 30 lbs. that the mass would not have weighed any- 


——— thing under water, but would have been 
— 6 lbs. able to buoy up 6 lbs. 


ProsLEM 3. A ship started at the equator and sailed 3 
degrees north, then 2 south, then 5 north, and then 9 south. 
In what latitude was it then? 








OPERATION. 
3 degrees, 
— 2 degrees, EXPLANATION. Consider north lati- 
5 degrees, tude positive. Then south latitude must 
— 9 degrees, be negative. The above course ends 3 
7 degrees south of the equator. 
— 3 degrees. 


Prosiem 4, If an ant running up a tree ascends 4 feet, 
then falls back 2 feet, then ascends 9 feet, and then falls 
back 5 feet, what progress has it made? Ans. 6 feet. 

Let the pupil explain it. 


28 ELEMENTARY ALGEBRA, 


SECTION VI. 


Problems to be solved by Equations containing 
but One Unknown Quantity. 


1. A FARMER had 5 times as many cows as horses, and 3_ 
times as many sheep as cows. The number of them all was 
105. How many of each kind had he? Ans. 5 horses, 25 
cows, and 75 sheep. 


2. From 64 times a certain number, if 36 times the num- 
ber be subtracted, 7 times the number added to the remain- 
der, and 12 times the number subtracted from the last 
result, the remainder is 46. What is the number? Ans. 2. 


3. Four boys have in all 84 cents. The second has three 
times as much as the first; the third has as many as the 
difference between what the first has and what the second 
has; and the fourth has as many as the difference between 
twice what the first has and what the second has. How 
much has each? Ans. The first has 12 cents; the second, 
36 cents; the third, 24 cents; and the fourth, 12 cents. 


4, Two men are 175 miles apart, and travel toward each 
other. One travels 15 miles a day, and the other 20. In 
how many days will they meet? Ans. 5 days. 

Notre. Let x stand for the number of days required. Then the 


first will go x times 15 miles, or 152, and the second will go z times 
20 miles. 


5. If 1362 —121a” = 150, what is the yvaluevor a 
Ye ee 


6. In a certain house there are twice as many doors as 
there are rooms, and twice as many windows as there are 
doors. There are 21 more windows than rooms. How 
many doors are in the house? Ans. 14 doors. 


7. There are 8 numbers, the second of which is four times 
the first, and the third is twice as much as the first and 


ELEMENTARY ALGEBRA, - 29 


second together. The difference between the second and 
third is 86. What are the numbers? Ans. 6, 24, and 60. 
8. Two men are 42 miles apart, and travel toward each 
other. One goes 3 miles an hour, and the other 4. In how 
many hours will they meet? Ans. 6 hours. 
9. Given, « + 16x — 38az = 42, to find the value of z. 
Ans ane +3, 


10. A bankrupt owed A 4 times as much as he owed B. 
He owed C 3 times as much as he owed A. He owed D the 
difference between his debts to C and B. The amount of 
his debts to all four was $28000. How much did he owe 
D? Ans. $11000. 


11. A man bought 3 horses and 4 cows for $600. Each 
horse cost twice as much as each cow. What did he give 
for each? Ans. He gave $120 for each horse, and $60 for 
each cow. 


Notre. Let z represent the price of a cow; 22, of a horse; 42, 
of all the cows; and 62, of all the horses. 


12. James bought 16 apples at one time and 5 at another, 
all at the same rate. The first time he paid 33 cents more 
than the second. What was the price per apple? Ans. 3 
cents. 

13. Seven men and three boys were hired for a week, 
each man receiving 3 times as much as each boy. Altogether 
their wages amounted to $72. What did each receive? 
Ans. Each boy received $3, and each man received $9. 

14. A farmer sold some corn, rye, and wheat; receiving 
3 times as much a bushel for wheat as for corn, and twice 
as much for rye as for corn. There were 20 bushels of 
wheat, 16 bushels of rye, and 40 bushels of corn. He re- 
ceived in all $66. What was the price per bushel of each 
kind of grain? Ans. He received $1.50 a bushel for the 
wheat, $1.00 a bushel for the rye, and 50 cents a bushel for 


the corn. . 
8 % 


30 ELEMENTARY ALGEBRA. 


Find the value of 2 in the following equations. 
15. 852 + 17x — 822 4 384—16x2 = 16 —7 4 19. 


Ans. & = 4, 
16.2+ 22%—8274+42+4+ 54—62+ 72x = 130. 
Ans. x = 18. 
WW.24+2r4+ 38a —4xe24+ 5¢ —6e2=74+84 12 
— 17. Ans. x = 10. 
18.¢@—Ta#+ 8e+4+1l2=7 + 35 + 62 — 24. 
Ans. x = 10. 
19. 2642 —14274+ 32 —1llaxz = 36 + 42 + 12 — 30 
+ 8. Ans. © = 17. 


20. 88a —62 + 262 = 37 — 244+ 103. Ans. x = 2, 


SECTION VIL. 
Subtraction. 


PROBLEM 1. Subtract 42 from 72. 


OPERATION. 
(har EXPLANATION. It is evident that the dif- 
4g. ference is 3x. This is the same as if we had 


—— changed the sign of the quantity to be subtracted, 
3m. making it — 4x, and then added it to 7x. 


PROBLEM 2. Subtract 52 from 22. 


OPERATION. EXPLANATION. It is evident that after 
es taking from 2 7 as much as we can of the 52, 

5 2, 3% will remain to be subtracted. That is what 

is meant by —3wa. This is the same as if we 
— 32. had changed the sign of the quantity to be sub- 
tracted, making it — 5a, and then added it 





to 22x. 


ELEMENTARY ALGEBRA. ol 


PROBLEM 3. Subtract — 2x from 642. 


OPERATION. EXPLANATION. It is evident that 62 is 
6 x, the same as 8x—2z2. If we subtract or take 
— 22, away the —2., the 8x remains. This is the 





same as if we had changed the sign of the quan- 
8 x. tity to be subtracted, making it + 2, and then 
added it to 6x. 


ProstEem 4. Subtract — 3-2 from — 8x. 


OPERATION. EXPLANATION. It is evident that —8-2x 
— 8a, is the same as —5x2—8zwx. If we subtract 
id a, or take away the — 32, the — 5a remains. 


This is the same as if we had changed the sign 
— 5x. of the quantity to be subtracted, making it + 32, 
and then added it to — 8a. 





PROBLEM 5. Subtract — 7x from — 2a. 


OPERATION. EXPLANATION. It is evident that — 2 z is 
— 2x, the same as 5&2 —7ax. If we subtract or take 
orl a away the — 72z,the 5x remains. This is the 


same as if we had changed the sign of the sub- 
5 x. trahend, making it + 7 x, and then added it to 
the minuend, — 2 x. 





PrRoBuLeEM 6. Subtract } from ec. 


OPERATION. 
c EXPLANATION. It is evident that the re- 
b, mainder is c— 06. This is the same as if we 
had changed the sign of the subtrahend, making 
e— b. it — b, and then added it to the minuend, ¢. 





Adding anything with a + sign, and the same thing 
with a — sign, does not alter the value of a quantity. 


Thus: a is the same asa + x—wZ. 
5c is the same as 5c + 36— 8380. 


oe ELEMENTARY ALGEBRA. 


PROBLEM 7. Subtract — 6 from ec. 


OPERATION. EXPLANATION. It is evident that ¢ is the 
c same asc + 6 —b, since + b and —b amount 
— b, to nothing. Subtracting or taking away the 
—b, the remainder is ec + 6. This is the 
e+ b. same as if we had changed the sign of the sub- 
trahend, making it + 6, and then added it to 

the minuend, c. 





To subtract anything is to take it away. It cannot be 
taken away from anything unless it is there. But it can 
always be got there by adding it to the other with both a + 
sign and a — sign; for the two balance each other and 
amount to nothing, as + 6 — bin the example above. If 
then we take it away, the same thing with the opposite sign 
will be left added to the minuend. 

Thus we see that the operation of subtracting is just the 
same as changing the sign of the quantity to be subtracted 
and adding it to the minuend. Similar terms should be set 
down in the same column. It is best not to actwally change 
the sign, but only to imagine it changed, and then set down 
the result of adding it to the minuend. 


ProsueM 8. Take 3a + 5c — 2d — 2y from 7a — y 
+ 2¢— 7d. 


ProspuEM 9. Take 5ax — 2y + c from 38ax — 2y. 
PropiEM 10. Take 4cx + ca? from dex + c?x? + 3 cx’. 
a 11. Take 8ab + 7a) + x from 3a’) — =z. 
PRoBLEM 12. Subtract 9 ab — 3 ab? from 5 ab? — 6 a’b’, 


PRoBLEM 13. Subtract 21 y — 4 from 3 y* — 2. 


aOaNIaark WN 


ht pe 
= Oo © 


a a ee ee 
Oo © onal & Oo kP Cb 


bo 
hd 


ELEMENTARY ALGEBRA. oa 


OPERATIONS. 

8. Ta— yt 2c— 7d, 9. 8au—2y 
3a—2y + dc — 2d, Sax —2y+¢, 
4a+t+ y—3e— dd. — 2ax —¢. 

10. 5er + cx? + 3 cz2’, 11. 3a’7b — x, 
4cx + 2°’ 8ab+7vb+4+ a, 

Cx + 3cx’, — 8ab — 4a°b — 2x. 

12, 5ab?—6a° 13, 8yt—2 

—3ab? + 9a’), —4+ 21y, 
an 00) — 9 ab: 38y4+2—21y. 
EXAMPLES. 

. Subtract 8 x from 92. Ans. x. 

. Subtract 167 + 3y from19xz+ 5y. Ans. 8x 4 2Qy. 

. Subtract 18 z — 10 from 14z— 12. Ans. —4z— 2. 

. Subtract 16 y — 15 from 17 y + 38. Ans. y + 18. 

. Subtract a from a + zx. Ans. x. 

. Subtract 2 — 10 from z. Ans. 10. 

. Subtract x — 15 from 15. Ans. — x + 380. 

. Subtract a + 6 + ¢ from a — 6 —e. 


Ans. —26—2e. 


. Subtract a —b—cfroma+6+e Ans. 26 + 2e. 
. Subtract — 105 from 21. Ans. 126. 
. Subtract 35 from — 4. Ans. — 39. 
. Subtract a from — 6. Ans. — a — b. 
. Subtract 6 from — 6. Ange 2h: 
. Subtract — b from b. Ans. 2. 
. Subtract 7 # from — 22. Ans. — 9x. 
. Subtract — 5a from 122. Ans. 17 x. 
. Subtract 10 from 5. Ans. '— 5; 
. Subtract 17 from 6. Ans; = 11; 
. Subtract 29a from 14a. Ans. — 15a. 
. Subtract 87 x from 27 x. Ans. — 102 
. Subtract — 15 from 5. Ans. 20. 


34 


22. 
23. 
24. 
20. 
26. 
27. 
28. 


29. 
30. 


ol. 
32. 


9 


3. 
34. 
30. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43, 


44, 


ELEMENTARY ALGEBRA, 


Subtract — 15 from — 5. Ans. 10. 
Subtract 15 from 5. Ans. —.10. 
Subtract 15 from — 5. Ans. — 20. 
Subtract x + y from x7 + y. Ans. 0. 
Subtract 63 from — 63. Ans. — 126. 
Subtract — 6 from 6. Ans. 12. 
Subtract 23 a’a’y from 5 a’a’y. 


Ans. 5 a’a*y — 23 a7aty, 
Subtract 16 amn from 5 amn. Ans. —11amn. 
Subtract 144 abex from 25 abs. ' 
Ans. 25 ab2 — 144 abex. 
Subtract x from y. Ans. y —&. 
Subtract 12 2 — 3 from 138 y — 4. 
Ans. 134 —12¢% —1, 
Subtract 6z + 43 from 122 — 21. Ans. 6z — 64. 
Subtract 11 a —2¢ from 9a —5e Ans. —2a—3e. 
Subtract 15 2 —13 y — 2z from 17x —13y 4 z. 
Ans. 2% + 32. 
Subtract 13a —26 + & from 23a + 116 —A. 
Ans. 10a + 136—A—kE. 
Subtract —15a’a + Tay —15 from 21 a’a% — 18 a’y 
+ 2. Ans. 36 Wa —20a?y + 17. 
Subtract 192z + 38yz — 14x22’ from 138 #z —12yz + 15. 
Ans. —6az—1dyz + 1422? + 15. 
Subtract 2m’n — 5 hn + 6 from 25 m’n —14 mn? + ha 


— 3. Ans. 23 m’n —14 mn? + 6hn — 9. 
Subtract 13¢@+ 36 + 5c from 18a—306 + 5e—1. 
Ans. —6b —1. 


Subtract 129 a? + 12 a@’x —ay + ax from 189 a? — 36 
a'r + ay—3. Ans. 60a? — 48 a’x + 2 ay — ax —3. 
Subtract 13 azy — 2b —19cx —5 from 17axy —136 


1c — 2. Ans. 4axy —11b+ 8lez 4+ 3. 
Subtract c + « + y —7b from 23mn —7b + e—a 
—y. Ans. 23mn —22”—2y. 


Subtract 5 a? —9 pq + 3px —z from 13 pq — 2px + 
a? —z, Ans. 22 pq — 5d px —4 a’. 


45. 
46, 
47. 
48. 
49, 
50. 


51. 


52, 
53. 
54. 
5D. 
56. 
57. 
58, 
59, 


60. 


ELEMENTARY ALGEBRA. 35 


Subtract az’ + 10 ax — 20a from 20 az? — 10 ax + a. 
Ans. 19 ax? —20 ax + 21a. 
Subtract 217 amv —18w—3u —5D from 193 amv — 


18w + 5u —6. Ans. — 24amv + 8u—1. 
Subtract 22? —16q — 28m + «'*y’ from 15 a*y? — 29 
m + q. Ans. 14 ay’ —m + 17q —22’, 


Subtract 6 a? —5a + 21 from 2 a?. 
Ans. —4a? + 5a —21. 
Subtract —5hk + 3k? — 36 from 3hk — 21. 
Ans. 8hk —3k? + 15. 
Subtract 9 a? from — 13 a@—3a + 6. 
Aus. — 220 —s3a + 6; 
Subtract 172 2° + 102°y —3ay’ + y* from 250 «® — 
120 xy + dry’ — y’. 
Ans. 78 x —1380a’%y + 82y’? —2 y’. 
Subtract 1385 c’d + 6my —a’x from 170¢’d —a’x + 
5 my. Ans. 35¢’d — my. 
Subtract 7 aw —10 by from 13 ax + 5 by. 
Ans. 6ax + 15 by. 
Subtract —5 ab + a? —0’ from 26 ab. 
Ans. 31 ab —a? + 0’. 
Subtract 7 ab? —5a’y + 3a’b from 2a°b — 5a’y + 


3 ab’. Ans. —a’b —4 ab 
Subtract « + y and —a + y from x —y. 

Ans. « —3y. 
Subtract 9a —b, —Ta + b, and 6a + 36 from 5a 
—3b.. é Ans. —3a —6b. 


Subtract « + y +z and —xz —y +2 from x —y —z. 
Ans. x —y —32z. 
Subtract Tay —3ax + ay from 9x —3ay + 2x —5bday. 
Ans. 144” —16ay. 
Subtract 8 aty? —6a'y + 3 a‘y’ from 13 ay —9 aty’ 
— bay. Ans. 14 aby — 20 aty’. 


Norr. Examples in Subtraction may be proved by adding the 
subtrahend to the remainder. The result should be the same as the 
minuend, 


36 ELEMENTARY ALGEBRA. 


SECTION VIII. 
Transposition. 


In the equations previously given, all the unknown quan- 
tities were on one side of the sign =, and all the known 
quantities werc on the other. For example, we had the 
equation, 

30a” - 17x—382a"+ 34—162 = 16—7 419. 
Here, in the first member of the equation, every term con- 
tains x; and in the second member, every term is known. 
That being the case, we may perform on them the operations 
which their signs indicate, and obtain as a result, 


(ae eos. 
Then, dividing both members by 7, 
co 4) 


It is sometimes the case that equations do not contain the 
unknown quantities thus separated from the known ones by 
the sign of equality. Then we must separate them, or the 
equations cannot be solved. For example, suppose 


5a2—4= 6. 


Now, if by taking 4 away from 5 we get 6, the 5x must be 
4 more than 6. That is, 


d”a2=6-+ 4. 


The equation can now be solved by collecting the terms and 
dividing : 
ee ANEE 
eae ek 


It will be seen that whereas — 4 was in the left member of 
the equation at first, + 4 is in the right member of the new 
equation. This operation is called transposition; and it will 
be shown that any quantity may be removed from one mem: 
ber of an equation to the other, if its sign be also changed. 


ELEMENTARY ALGEBRA. 2 yi 


PROBLEM 1. What is the value of x in the equation 7 x 
—9 = 26? 

SoLution. Here we wish to transpose —9. We can do 
it by adding + 9 to both members of the equation. 

(zt —9 = 26 
+—+- 9== + 9, 
7x = 26 + 9. 

Thus we have obtained a new equation. The — 9 disap- 


pears, and + 9 appears on the other side of the equation. 
Solving now this new equation, 








7x2 = 26 + 9. 
Tx = 35, 
x= 5, 


ProBLEM 2. Given 8x2 = 24+ 52, to find the value 
of x. 


Sotution. Here we wish to transpose + 5a”. We can 
do it by adding — 5 to both members of the equation. 
8x = 24+ 52. 


—b)2x= — 5a, 


82—bH x2 = 24. 
In the new equation, the + 52 disappears, and — dz 
appears on the other side of the equation. Solving now this 
new equation, 


ra $. 


It will be observed, that to transpose a quantity we add 
the same quantity with a contrary sign. 


PROBLEM 3. Solve the equation 7x —14 + 3x45 
aol. 
SoLuTion. Here we wish to transpose the — 14 and the 


+ 5, so as to leave the terms containing x by themselves. 
4 


38 ELEMENTARY ALGEBRA. 


We can do it by adding these quantities with their signs 
changed to both members of the equation. Thus, 


Tae—14+32+4+5= 31. 








(fy +32 =314+14—5. | 
Collecting, 102 = ae 
Whence, c= 4, 


Now it is plain that adding quantities with their signs 
changed to that side of an equation in which they are, will 
always remove them from that side, since the same thing with 
both a + sign and a — sign amounts to nothing. It is 
also plain that doing the same thing to the other side of the 
equation will always introduce those quantities with their signs 
changed to that other side. And adding the same thing to 
the equal members must give equal results, thus giving us 
a new equation. 


PrRoBLEM 4. Given 7x = 63 —2z, to find the value 
of x. 


Sotution. Add 2.2 to both members, and we have 





(2x = 635 —22. 
+2z2=> + 2a 
Tx + 2x = 63. 
Collecting, 9a = 63. 
Dividing by 9, co =e 


ProsiEM 5. Given 132 —2 = 4a -+ 48, to find the 
value of x. 


SoLution. Add + 2 and — 42 to both members. 





13 2—2 = Ax 43. 
+2—4%4=—42 + 2. 
Collecting, Ga = 46. 


Dividing by 9, Ferma 


ELEMENTARY ALGEBRA. 39 


It is not customary to set down all this operation of ad- 
dition; but, since we know what the result must be, to set 
. that down at once, as will be done in the following examples. 
We must be careful not to leave out any term. It is best 
first to set down on each side of the sign of equality the 
terms which are to remain on that side, then those which 
are transposed to it from the other side. 


PrositEM 6. Given 1227 —14+4+ 8x2 = 72 4+ 52 +4 4, 
to find the value of z. 


SOLUTION. 
Transposing, 127 + 382—5H%=72+4+4+4 14. 
Collecting, 102 = 0. 
Dividing by 10, Cn 
ProsieM 7. Given —2x% + 41+7=>—2+8-—2 
+ 14, to find the value of x. 
SOLUTION. 
Transposing —2r+er=>—24 84 14—A41 —7. 
Collecting, — x2 = — 28. 
And if that is the case,z = 28. 
EXAMPLES. 
Find the value of « in each of the following equations: 
foe tO — 2.0 = 6. Ans. « = 1. 
2.52 +11 —2a2 = 20. Ans. x = 38. 
3. 4x —6 + 5 = 33 —14. Ans. 2 = 9. 
4.12+ 5x2 = 108 —5d52 —6. Ans. x = 9. 
5. 14= 60+ 4x2 —8a2 + 2. Ans. x = 12. 
6. ¢= 56 —8 — 3m. Ans, 0 212, 
72xn+7=24108. Ans. « = 8. 
8. 26 —4"% =2x+5—3 —2e. Ans. x = 8. 
9.144 32 = 78 —z. : Ans. x = 16. 
10. 32 —30 = 41 —2 + 2-. Ans. x = 69. 
11. 88 —12 = 14 + 9x2 — 5d. Ans..%@ =+13 


40 


12. 
13. 
14. 
15. 


16. 
ive 
18. 
19. 
20. 


ELEMENTARY ALGEBRA. 


3x2 —26 = 39 —8. Ans. « = 19, 
39 + 2a = 4x —A?7. Ans. x = 48. 
27 = 50 —724 61. Ans. x = 12. 
8x = 14x” 4+ 14 —55 — 8b. Ans. « = 21. 


Find the value of y or z in the following equations: 
22 —6y + 18 = 100 —14y 4 86. Ans, Y= ie 


sy—6=y+ 16 —6. Anas, Yi=ae 
6+ 2z2=> 5z— 36 + 21. Ans. z = 7. 
41 —6z2—17 = 120 —14z. Ans. z = 12. 
82 —13z = 92 —15z. Ans. z= 86. 


Notr. Some equations may be somewhat simplified by merely 
omitting something which occurs in both members. Thus, in the 
17th of the preceding examples, we may subtract the —6 from both 
members; that is, we may take it away or omit it, leaving the 


equation 


By =y+ 16. 


Then the transposition and the rest of the solution is somewhat 
simpler. 


SECTION IX. 


Questions producing Equations that require 


Transposition. 


PRoBLEM 1. What number is that which with 7 added 
to it amounts to 23? 


SoLution. Let x represent the number. 


Then we have the equation, e+ 7 = 23. 
Transposing, x = 23 —T. 
Collecting, Das be 


PropiEeM 2. Three partners, A, B, and C, have a capital 
of $56000. B’s is $5000 more than A’s, and C’s is $6000 
iess than A’s. What is the capital of each? 


ELEMENTARY ALGEBRA. 41 


Sotution. Let x represent A’s capital. 
Then x + 5000 represents B’s. 
And «x — 6000 represents C’s. 
We have the equation 
x + x + 5000 + « — 6000 = 56000. 
Transposing, «+ « + « = 56000 + 6000 — 5000. 


Collecting, 32 = 57000. 
Dividing by 3, zie 19000. 
Whence, xz + 5000 = 24000. 
And x — 6000 = 13000. 


Therefore A’s capital is $19000; B’s, $24000; and C’s, 
$13000. 


ProBLEM 3. B bought two horses and a $50 harness. 
The first horse was worth the second horse plus the harness. 
Twice the value of the second was equal to the value of the 
first plus the harness. What was the value of each horse? 


Sotution. Let x stand for the value of the second horse. 
Then x + 50 stands for the value of the first horse. 


We have the equation Qx2—=>a2+ 50+ 50. 
Transposing,  ° 22 —x = 50 + 50. 
Collecting, 2 = 100. 

Whence, zx + 50 = 150. 


Therefore the first horse was worth $150; and the second, 
$100. 


EXAMPLES. 


1. From three times a certain number 5 is subtracted, and 
the remainder is 13. Whatisthe number? Ans. 6. 


2. A pole 23 feet long is stuck up in the middle of a pond, 
going into the mud at the bottom 2 feet. Twice the length 
in the water multiplied by 3 is the length in the air. How 
much is in the air? Ans. 18 feet. 

3. The difference between two numbers is 1, and their 


sum is 13. What are the numbers? Ans. 7 and 6. 
4% 


42 ELEMENTARY ALGEBRA. 


4, A man worked three days for $14. The second day he 
earned $1 more than. the first; and the third day, as much 
as on both the first and the second. How much did he earn 
each day? Ans. He earned $3 the first day, $4 the second 
day, and $7 the third day. 

5. I met a man driving a flock of geese, and said, ‘‘Good 
morning, master, with your hundred geese.” He replied, 
“T have not a hundred; but if I had twice as many as I 
now have and 4 geese more, I would have a hundred.” If 
he spoke the truth, how many had he? Ans. 48 geese. 


6. George is 5 times as old as John. The difference of 
their ages plus 24 years is twice the sum of their ages. 
What is the age of each? Ans. George’s age is 15 years, 
and John’s is 3 years. 


7. James has 5 times as much money as George, and 
William has 38 times as much as George. James has 22 
cents more than William. How much money has each? 
Ans. James has 55 cents; William; 33; and George, 11. 

8. A and B entered into partnership, A contributing 3 ~ 
times as much to the stock as B. The sum of their stocks 
is worth the difference of their stocks plus $8000. What 
did each contribute? Ans. A contributed $12000; and B, 
$4000. 

9. The head of a fish is twice as heavy as the tail; the 
body weighs 2 pounds more than twice the weight of the 
head and tail together ; and the whole fish weighs 20 pounds. 
What is the weight of the body? Ans. 14 pounds. 

10. The head of the same fish is 6 inches long ; its tail is 
half as long as its body; and its tail and body together are 
3 inches more than 4 times the length of the head. What 
is the length of its body? Ans. 18 inches. 

11. John bought a certain number of tops, and 15 times 
as many marbles. After losing 14 of his marbles, and giv- 
ing away 30, he had only 16 marbles left. How many tops 
did he buy? Ans. 4 tops. ; 


ELEMENTARY ALGEBRA. 43 


12. A and B can together earn $40 a week, of which B 
can earn $4 more than A. What can each earn? Ans. A 
can earn $18 a week; and B, $22. 

13. There are 38 more counties in Connecticut than in 
Rhode Island, and 6 more in Massachusetts than in Connec- 
ticut. In the three States there are 2 more than 5 times as 
many as in Rhode Island. How many are there in each? 
Ans. In Rhode Island, 5; in Connecticut, 8; and in Mas- 
sachusetts, 14. 

14. Divide the number 43 into two parts, one of which 
shall be 15 more than the other. Ans. The parts are 29 
and 14. 

15. Divide the number 43 into two such parts that 3 
times the less part shall be 5 more than the greater. Ans. 
The parts are 31 and 12. 

16. Divide the number 48 into two such parts that twice 
the less part shall be 4 less than the greater. Ans. The 
parts are 13 and 30. 

17. In my largest bookcase are 3 times as many books as 
in my smallest and 60 books besides. In another, twice as 
many as in the smallest and 20 books besides. In both, 6 
times as many as in the smallest. How many are in each? 
Ans. In the largest bookcase there are 300 books; in the 
smallest, 80; and in the other, 180. 

18. Three pieces of marble weigh in all 15 times as much 
as the smallest piece and 19 pounds besides. The second 
weighs 3 pounds more than 5 times the first, and the third 
twice as much as the second. What does each weigh? 
Ans. The first piece weighs 10 pounds; the second, 55 
pounds; and the third, 106 pounds. 


19. What number is that which will be doubled by add- 
ing 5 to it and subtracting the sum from 41? Ans. 12. 


20. Divide $47 among George, John, and Charles, so that 
Charles may have $1 more than George, and John $3 more 


44 ELEMENTARY ALGEBRA. 


than Charles. Ans. George’s share is $14; John’s, $18; 
and Charles’s, $15. 

21. A person spends $24 for clothing, and $10 for books, 
and then has remaining one-third of what he had at first. 
What has he remaining? Ans. $17. 


22. There are four numbers whose sum is 18 more than 
twice the third. The second is 2 more than twice the first; 
the third is 3 times the first; and the fourth is 1 more than 
the third. What are the numbers? Ans. 5, 12,15, and 16. 

23. Three boys spend at a candy-shop 36 cents. The 
second spends 1 cent more than twice as much as the first ; 
and the third spends 4 times as much as the first. What 
does each spend? Ans. The first spends 5 cents ; the second, 
11 cents; and the third, 20 cents. 

24. A mother’s age added to her daughter’s gives 40 years, 
and the mother’s is 4 years more than 3 times the daughter’s. 
What is the age of each? Ans. The mother’s age is 31 
years, and the daughter’s is 9 years. 


25. At a certain election 2874 persons voted, and the suc- 
cessful candidate had a majority of 376. How many per- 
sons voted for each? Ans. 1625 persons voted for the suc- 
cessful candidate, and 1249 for the other. 

26. Divide the number 58 into three such parts that the 
first may exceed the second by 9 and may exceed the third 
by 17. Ans. The parts are 28, 19, and 11. 

27. Of $120, I spent a part. Three times the remainder 
are $4 more than I spent. How much did I spend? Ans. 
$89. 

28. A started to overtake B, who was 18 miles ahead of 
him. Both were on horseback, and A rode 10 miles an hour 
while B rode but 8. How long was A in overtaking B? 
Ans. 9 hours. 

29. How can an estate of $12000 be divided among a 
widow, her son, and her daughter, so that the widow shall 


ELEMENTARY ALGEBRA. 45 


have $2000 more than her daughter, and $2000 less than 
her son? Ans. The widow’s share is $4000, the son’s is 
$6000, and the daughter’s is $2000. 

30. If the difference between 4 y + 20 and 2y —16 is 
equal to the difference between 84 —3y and 12 — y, what 
is the value of y? Ans. y = 


SECTION. X. 
Multiplication. The Signs + and —. 


Mourtiety + 3 by + 2. This means that + 3 is to be 
multiplied by 2, and the result added. Hence, the product 
is + 6 added, which is + 6. 

Multiply —3 by + 2. This means that — 3 is to be 
multiplied by 2, and the result added. Hence, the product 
is — 6 added, which is — 6. 

Multiply + 3 by —2. This means that + 3 is to be 
multiplied by 2, and the result subtracted. Hence, the pro- 
duct is + 6 subtracted, which is — 6. 


Multiply —3 by —2. This means that —3 is to be 
multiplied by 2, and the result subtracted. Hence, the pro- 
duct is — 6 subtracted, which is + 6. 


Thus we have had: 
+ 38 multiplied by + 2 = + 6, 
— 3 multiplied by + 2 = — 6, 
+ 3 multiplied by — 2 = — 6, 
and — 3 multiplied by — 2 = + 6. 

In these cases, when the signs of the two factors are the 
same, the sign of the product is +; and when the signs of 
the two factors are different, the sign of the product is —. 

Now this must always be so. To multiply any + quan 


46 ELEMENTARY ALGEBRA, 


tity by a + quantity is simply to take it so many times and 
add the result; and to multiply any — quantity by a — 
quantity is to take it so many times and subtract the result. 
In both cases the product is +. Also, to multiply any — 
quantity by a -++ quantity is to take it so many times and 
add the result; and to multiply any + quantity by a — 
quantity is to take it so many times and subtract the result. 
In both cases the product is —. 

Hence, whenever we multiply two quantities together, like 
signs give +, and unlike signs give —. 


EXAMPLES. 
1. Multiply + 8 by — 2. Ans. — 16. 
2. Multiply — 8 by + 2. Ans. — 16, 
3. Multiply — 8 by — 2. Ans. 16. 
4. Multiply + 8 by + 2. Ans. 16. 
d. Multiply — 13 by + 21. Ans. — 273. 
6. Multiply — 18 by — 5. Ans. 90. 
7. Multiply 6 by 7. Ans. 42. 
8. Multiply — 6 by 7. Ans. — 472, 
9. Multiply 6 by — 7. Ans. — 42. 
10. Multiply — 5 by — 18. Ans. 65. 
11. Multiply 6 by a. Ans. 6 x. 
12. Multiply 6 by — a. Ans. — 62. 
13. Multiply — 6 by a. Ans. — 62. 
14, Multiply — 6 by — 2a. Ans. 6 x. 
15. Multiply —11 by 22. Ans. — 22 x. 
16. Multiply 4 by —7a. Ans. — 28%. 
17. Multiply — 5 by — 3, Ans. 15 x. 
18. Multiply 4 by —a. Ans. — 4a. 
19. Multiply —2 by —e. Ans. 2c. 
20. Multiply — 5 by 3. Ans. —15¢, 


ELEMENTARY ALGEBRA. 47 


SECTION XI. 
Multiplication of Literal Quantities. 


WHEN we write two letters together without any sign 
between them, we mean that one is to be multiplied by the 
other. Thus, ad means the product of a and b. So does ba. 
In the same way, xyz means the product of «, y, and z. 
This is not the case with figures. Thus, 53 does not mean 
5 times 3, but 5 tens added to 3. A figure before a letter, 
however, indicates multiplication: 4n means 4 times n. 


EXAMPLES, 
1. Multiply 9 ad by 6c. Ans. 54 abe. 
2. Multiply 5ay by —n. Ans. — dnxy. 
3. Multiply — a’ by — 8’. Ans. a’b’. 
4. Multiply — 8a’x by 8 ny. Ans. — 24 a’xny. 
5. Multiply — 15 yz by —3 a”, Ans. 45 yzx’. 
6. Multiply 6 a? by — 82 ct. Ans. — 192 a’x*. 
7. Multiply — 42 abcdef by gha. Ans. — 42 abcdefghi. 
8. Multiply — 2 ca*y® by — 5 n’a. Ans. 10 cay*n?a. 
9. Multiply 22 mn by 3 p’g. Ans. 66 mnp’q. 
10. Multiply —4mn by — 2 az. Ans. 8 mnax. 
11. Multiply —9 a by 8y. Ans. —72 xy. 


12. Multiply —8z by —9y. Ans. 72 xy. 
13. Multiply 4 cn by 4 a. Ans. 16 ena. 
14. Multiply 3 2’ by — y’. Ans. —3 x*y', 
15. Multiply —2y by 7 nh. Ans. —14 ynh. 
16. Multiply — 11 xz by — 6 ws. Ans. 66 xzus. 


hoe ee 
Gq <o Go =] 


. Multiply 3 n’ by 15 p’q’. 

. Multiply —2 zy by — 10 zw. 
. Multiply 4 acx by — 3 pq. 

. Multiply ——3 nae by 7 ad. 


Ans. 45 n’p*¢’. 
Ans, 20 xyzw. 
Ans. —12aexpgq. 
Ans. — 21 nxcab. 


48 ELEMENTARY ALGEBRA, 


SECTION XII. 
Multiplication. Exponents. 


Ir we multiply a by a, we may write the product aa; but 
it is more usual to write it a. So, if we multiply together 
x, x, xX, z,and x, we may write the product axvxxx; but we 
usually write it z*°. The numbers placed to the right of the 
upper part of a letter are called exponents or indices. Their 
use is to show how many times the letter is to be taken as a 
factor. The letter x alone is the same as =’. 

Now, if we multiply aaa by aaaa, the product is aaaaaaa. 
That is, a? x a* = a’. The exponent of a letter in the 
product is equal to the sum of its exponents in the factors. 
This must be so; for the letter is taken as many times in 
the product as in the factors, and the exponent only shows 
how many times it is taken. 

Hence, ce xX c' = ¢;‘for ce X ¢ =. cee. 
nxX n= n'; forn X nnnnn = nnnnnn. 
eX 8 =e a0 OT cee ok eens eo ee 

In multiplying one term by another, remember: 

1. Like signs give +; unlike signs give —. 

2. Writing letters together expresses their product. 

3, The index of a letter in the product is the sum of its in- 
dices in the factors. 


EXAMPLES. 
1. Multiply a by 8. Ans. ab. 
2. What is the value ofa x bX ¢ X x? Ans. abex. 
3. Multiply 3 a by 7 a. Ans. 21 ax. 
4. Multiply 122 by 9 y. Ans. 108 xy. 
5. Multiply — 6a by 50. Ans. — 30 ab. 
6. Multiply 5axy by —7 bz. Ans. — 35 absryz. 
7. Multiply 9 aaaaa by 5 aaa. Ans. 45 a®, 
8. Multiply 21 aabbb by 32 abbce. Ans. 672 a%b'c’, 


ELEMENTARY ALGEBRA. 49 


9. Multiply 15 exyyy by 3 xxzxy. Ans. 45 x>y*, 
10. Multiply 9a° by 5 a’. Ans. 45 a’, 
11. Multiply 21 ab? by 32 ab’c’, Ans. 672 a5b'c’. 
12. Multiply 15 2’y° by 3 xy. Ans. 45 ay. 
13. Multiply 18 x*y® by 4 ay4z”, Ans. 52 xy'2’, 
14. Multiply 12 a*b’c’ by 4 a’ctrt, Ans. 48 a®b’c'x', 
15. Multiply 5 a®x’y? by — 6 aba, Ans. — 30 a®a® yz. 
16. Multiply — 6 ay*z* by — 4 23732’, Ans. 24 x®y®z*, 
17. Multiply — 5 xty’z by 7 a®y’. Ans. — 35 a'aty*z. 
18. Multiply 21 az*y” by 3a’xy’. Ans. 63 ax?y”. 
19. Multiply — 14 a®x%2? by 2axyz. Ans. — 28 a®xy*z*. 


. Multiply —4 2/52" by — 5 ayiz’. Ana 20 sfy 22". 
. Multiply — 38 atatz’ by 9 a®btx®yz. 


Ans. — 27 a®btx® yz’, 


. Multiply —7 axry'z by — 38 ary'z. Ans. 21 ax’y%2?, 
. Multiply — 4 a*m'nt by —9 a’min®=. Ans. 36 am?n™. 
. Multiply 12 ap*¢’rs by 3 abpqrs’. Ans. 36 a®p'q’r’s”. 
. Multiply 16 a’m®x%y by 12 atm’y'. Ans. 192 a®m'x®y’. 
. Multiply 5 a®m®x*z* by — 3 am’a%z*. 


Ans. — 15 a®m" 2725. 


. Multiply —0 @ax’y*z by 6 xty'2’. Ans. 0. 
. Multiply — 6 aty’z" by — 2 a®atz”, Ans, 12 a®aty®2?. 
. Multiply together 12 a 


"2°, — 5 a?z, and 6 a’. 


Ans. — 360 azz’, 


. Multiply together 5 m’n7p’q°, —3 m'n’, and — 7 p*q’. 


Ans. 105 mn*p'q™. 


. Multiply together — 12 a®b’xty’, —5atb’xty’, and — 2 


abr y*, Ans, — 120 a*b'x'ty’. 


. Multiply together 18 aSx’y, 3 a°z’, and — 2 ata®yz’. 


Ans. .— 108 a®x"y?z', 


. Multiply together 12 a®b‘c’d’e> and — 9 a’btce*. 


Ans. — 108 ab®c'd’e’. 


. Multiply together 11 a®’x’, — 15 az*y*, and — 14 ax’y*z’. 


Ans. 2310 a" ax'y*2’, 


. Multiply together — 3 a‘z*, — 17 az‘, and 5 ata*y. 


Ans, 255 a®x"y, 
5. D 


50 


36. 


ELEMENTARY ALGEBRA. 


Multiply together — 12 wx°z, —4 a®xz’, and — 3 axty®z. 
Ans. — 144 a¥aty®2". 


. Multiply together — 4 atz’, 6 ax’y%z’, and 7 2’. 


‘Ans. — 168 ax?y*z®, 


. Multiply together 12 m’n’p’, 4 m'n*p*, and 2 mnp. 


Ans. 96 m'n'p', 


. Multiply together a’, 12 a‘, 3 a’, a®, and a”. 


Ans. 36 a, 


. Multiply together a*x*, 5 a’x*, 3 aa*, 2 2, and a’. 


Ans. 30 a5z'8, 


. Multiply together 5 2? and atz’. Ans. 5 atx’. 
. Multiply together 0 a'a? and 5 ata’. Ans. 0. 
. Multiply together 7 a’z* and 13 a®y*. Ans, 91 aay’. 


. Multiply together 15 a‘v'y’ and 4 a®x®y‘z”. 


Ans. 60 a®a'®y®2?, 


. Multiply together — 7 a®aty’? and — 11 a*b'cta®y*2?, 


Ans. 77 aM beta y$2?, 
Multiply together — 2 a*e’u'd® and — 5 2°7*d¥a’. 
Ans. 10 a®e®d®xty?, 


. Multiply together — 18 a°xb*, — 2 b’ctx’, and — 8 2°b'a?, 


Ans. — 208 ab'cta™. 


. Multiply together — 5 atz?m®, 3 m*zx’a*, and — 2 a*a’m’. 


Ans. 30 a?m'*2?, 


. Multiply together — m,—m, —m, —m, —m, and —m. 


Ans. m*, 


. Multiply together — m, m, — m, m, —m, and m. 


Ans. —m*, 


. Multiply together — ax, ax, —ax, ax, —ax, and az, 


Ans. — a’x’, 


. Multiply together — ax, — ax, —- ax, — ax, and — aa. 


Ans. — abx', 


. Multiply together — az, ax, — ax, ax, and — ax. 


Ans. — aba’, 


. Multiply together 6 ab, — 2m’, — 9 ab, and 3 m’. 


Ans. 324 7b’m*, 
Multiply together 6 ab, — 2 m?, 9 ab, and 3 m?. 
Ans. — 324 a?b*m‘, 


_ ELEMENTARY ALGEBRA. 51 


56. Multiply together — 6 ab, —2 m’, — 9 ab, and 3 m’. 

Ans. — 324 a’b’m‘. 
57. Multiply together — 6 ab, — 2m’, —9 ab, and — 3m’. 
Ans. 324 07b’m‘, 

58. Multiply together — 6 ab, — 2m’, and — 9 ab. 
Ans. — 108 a’b’m’?. 

59. Multiply together 9 c’d’e*f*g*h' and — 2 c*h®f?dPe*. 

Ans. — 18 cdSe'f*g*h*. 

60. Multiply together — 2 atzx*c’y’ and — 5 &xty?a’. 
Ans. 10 a®c®x*y, 


SECTION XIII. 
Multiplication of a Polynomial by a Monomial. 


HitHERTO we have only multiplied together single terms. 
When we wish to multiply a quantity containing several 
terms, we must multiply all of its terms. To multiply y—z 
by z, we must multiply y by x and then —z by z. 


PrositEM 1. Multiply 7 ay—3a + 2¢ by —4 a’ex’, 


PrRoBLEM 2. Find the value of the following expression: 
(—9n? —3 cx* —1la’n) X nz. 


Nore. The parenthesis is necessary here. If it were omitted, 
the meaning would be that only —11 a?n was to be multiplied by 5 nz, 








OPERATIONS, 
+n 7 ay ee Oe + 2¢, 
— 4 acz’, 
— 28 atca’y + 12 a®cx*® — 8 ade?x?, 
va — 9n? — 8ext —I11a’n, 


5 nx, 





— 45 n'x — 15 ena® — 55 a’n?x. 


D2 


12 


18. 


ELEMENTARY ALGEBRA. 


EXAMPLES. 


Find the values of the following expressions: 


.(5a+ 24%—3y) X a. Ans. 5a? + 2ax —3 ay. 
. (8ax —3a’ + a’) X a’y. 


Ans. 8 aa®'y —38 aaty + sty. 


. (Tay —4a’y’ + Say’) x 8ay. 


Ans. 21 aty? —12 a®y® + 15 a’y*. 


. 18 aaty — 5 ax’y’? + Tay?) X —8 axy. 


Ans. — 54 a'aty’ + 15 a’aty’ OOF ax’y*. 


. (—9ata’y + da’ay’ —2y°) x —42°y*. 


Ans. 36 ataty® —12 aayt + 8 a’y'. 


. (5 ab — 207d? + 7 ab® —2b*) x —ab. 


Ans. —5 ath? + 2 a*°b? —7 a’b*t + 2ab. 


. (Gat — 5 ab? + 40°b? —7 abt) xX —8ab. 


Ans. —18a'b + 15 ath? —12 a°bt + 21 ad. 


. (—21 ming —14b2 + 162?) x —8 amie. 


Ans. 63 amma? + 42 a&bmtx? — 48 a®mtz3. 


. (12 aa — day’ — 3a’) x 8abaty. 


Ans. 36 a’a®'y — 15 a¥aty’ — 9 aay. 


. (10 a®xiz — Sarat’? + 7 aa®z? — 2a*‘xz) x 0 azz. 


Ans. 0. 


. (18 mn? — 11 ab + 5 ty’) xX Oatmiy. Ans. 0. 
. (abe — 8 ab + 6a —7) X —5Bae. 


Ans. —45a’be? + 40 a7?be — 30 a7e + 35 ae. 


. (18¢ —19 a + 16 2’y’) x 2a’. 


Ans. 36 a’ct — 38 ate? + 82 aeha’y’. 


. (16 ab —T7 bax + 9 a? — 8 ex) X Sean. 


Ans. 80 a°be?'x — 35 wbe'x? + 45 a&x — 40 ac®x’. 


. (12a —5a’b + 6a? —Qay) x —T aye. 


Ans. — 84atry + 35 a°®bxy —42a°xy + 14a’7ry’. 


. (19 ay? —13 min'a + 18y) x 5 xm. 


Ans. 95a&imz*y? — 65 &intntat + 90 may, 
(10 a3b’e —9 abe? + 38a°b'c? + 2a°bc®) x 8 abe. 
Ans. 80 aSdtc? — 72 atb’ + 24 abd’ + 16 abd8ct. 
(Saxy? —Tay + 6aa*y* —2ax) X —2axy. 
Ans. —10a@’a’*y’ + l4a’xy? —12 aaty? + 4a’a*y 


19. 


20. 


yA 


22. 


23. 


24, 


25. 


26. 


27. 


28. 


29. 


30. 


ELEMENTARY ALGEBRA. 53 


(3 e'd?a — 4 d’e’x? + 5 e’f?a*) x 6 efx’. 
Ans. 18 e’d’efx® — 24 d’e®fa* + 30 e*f*a’. 
(— 12 a®yz — 5 a?2’*x + 3 a'yz —Qayx) X — 3 ax*. 
Ans. 36 abx*yz + 15 atatz? — 9 aasyz + 6 abrrty, 
(— Dd axy’ + 2 a yx” —5 y’za — a’) X A aye. 
Ans. — 20 a’x’y®? + 8 ata’y? — 20 way?z — 4atvy. 
(—2aty® + 3a%m™) x 6 2%y%m’, 
Ans. —12 a m’ay® + 18 a®m7™ x97”, 
(5 A®k —60°k + 7 bh) x — 8 Dhk. 
7 Ans. — 40 bA°k? + 48 b&hk? — 56 b°A7k. 
(atpq — 7 pqx’ + 5 pq’x — x*p*q?) X 12 axqp*. 
Ans. 12 p'q?a' — 84 p'q’a> + 60 p'q?c*t — 12 p* qa. 
(14 b’e? — 8 a®b’ctard'e") x 4 ab cry. 
Ans. 56 a®b¥ce’xy — 32 a®b™ cde" a*y. 
(12 —5ctat + 2bex) x — abex. 
Ans. —12abex + 5 abbe’x — 2 ab’c*x”. 
(abe — be’a + bea? — abe?) X — abe’. 
Ans. — ab’ + a?’ — ath’? + atb’ct. 
(Sak —4ah + 3 —2k’) x 6ahk’. 
Ans. 30 ahk® — 24 ah?k® + 18 ahk® — 12 ahh’. 
(12 aca? —5ax’e’b + 4 x yb?) x Tatb’a. 
Ans. 84 a°b’cx’ — 35 abb’c’x® + 28 atbihx®y’. 
(5 atb* — 7 b*y* — 6 aty*) x —10 a°d%y’. 
Ans. —50 ab’y® + 70.a8b°y” + 60 ab y” 


SECTION XIV. 


Multiplication of a Polynomial by a Polynomial. 


To multiply anything by 3 + 5, we must not only take it 
3 times, but also 5 times. So, to multiply by x + y, we 
must first multiply by x, then by y, and add the products. 


ProstemM 1. Multiplya+odby ez + y. 


&* 


54 ELEMENTARY ALGEBRA. 


OPERATION. 

a + 8, 

co Y; 

ax + bx == the product of a + band z. 
+ ay + by, = the product of a + 5 and y. 





ax + bx + ay + by, = the product ofa+bandzty,. 


PRoBLEM 2. Multiply 30% + 4ca’y —3a°y' by 2c'm 
—8cx’y —6 xy’. 
OPERATION. 
8cea + 4eay— 3 2°y? 
20a — 8cecaxy— 6257 
6a’ + 8eay— 6 eaty’ 
— 24 &a®y — 32 eaxty’? + 24 cay? 

— 18 Caty’ — 24 ca®y® + 18 x8y4 

6 cx? — 16 ea*y — 56 erty’ + 18 xy‘. 





ProsiEM 3. Multiply 2® —2° + at —a2’+ 2?—2+41 
by 2 —a +1. 





OPERATION. 
®o@— O+ wt — P+ wo? — a2 +1 
xv—ae+l 
rmP— 2+ w— 2+ zt— vt 2 


Ee at ge — xs xt — ae + a? — xr 
a— gat gt— Gt 2?— atl 


ve —220' + 8a —38 24+ 8x —32' + 8a2’—224+1. 


In the last example, the multiplicand is said to be ar- 
ranged according to the powers of x, beginning with the 
highest. So is the multiplier. Hence the operation has a 
kind of regularity about it, and the product is also arranged 
according to the powers of z. In the example before it, the 
factors are arranged according to the powers of-e, beginning 


ELEMENTARY ALGEBRA. 55 


with the highest; and it so happens that they are also ar- 
ranged according to the powers of x and of y, beginning 
with the lowest. Such an arrangement is not always possi- 
ble, but should be made whenever it is possible. 


EXAMPLES. 
. Multiply « + y by x + y. Ans. x? + Q2ay + y’. 
. Multiply e+ d by e+ d. Ans. ’& + 2ed + d’. 
. Multiply 6+ «by b+ <x. Ans. 0? + 2bx + x’. 


. Multiply «+ ybya+ 0. Ans. ax + ay + ba + by. 
. Multiply m + n by p + q¢. 


Ans. mp + np + mq + nq. 


. Multiplya + bbye—d. Ans. ac + be —ad —bd. 
. Multiply A + kby x—y. Ans. hx + kx —hy — ky. 
. Multiply « + y by « — y. Ans. x? — y'. 
. Multiply ¢ + d by e—d. Ans. c? — d’. 
. Multiply ¢ + d by —e + d. Ans. —ec’ + d’ 
. Multiply r —s by r+ s. Ans. r? — 8”. 
. Multiply 2? + y? by «* — y”* Ans. x* — yt. 
. Multiply 2? + y by 2+ y. Ans. at 4+ 2a’y? + yf. 


. Multiply 2? —y’ by 2? —y*, Ans. at —22°y’ + yf. 
. Multiply a? + ax + a2 by a —=. Ans. a — x’, 
. Multiply a? + 2ax + 2’ by a —2ax + 2”. 


Ans. at —2a’?x? + x. 


; cage”. + 2ced+d’ bye + d. 


Ans. & + 8ed + 8cd’? + d, 


j Multiply e?—2ed +d bye + d. 


Ans. & — cd —cd’? + a’. 


. Multiply 2? —a2 +1 by «+ 1. Ans. x® + 1. 
. Multiply 2 —axz + ybyxa +a. 


Ans. x8 + ry —a’x + ay. 


. Multiply m? + m’n + mn’? + vn? by m —n. 


Ans. mt — n't. 


. Multiply 3m —22by3m+ 32. Ans. 9m? —42°, 
. Multiply 82* + 3y’ by 8a*—3y’. Ans. 642° —O9y*. 


56 ELEMENTARY ALGEBRA. 


24. Multiply 83—382+2?by3+ 2. Ans, 24 —~a# -+ 2%, 
25. Multiply a + at +a@+1by@—1. Ans. a? —1. 
26. Multiply c’ —é& + &—cbye’ +1. Ans. & —¢ 
27. Multiply 5 a®xry* — 8atx by 5 ary? + 3 ate. 
Ans. 25 a®x’y*t — 9 a®x’. 
28. Multiply p’? + q by p + ¢. 
Ans. p> + pq + pd + &. 
29. Multiply p®? + q@ by p + q. 
Ans. pt + p? + pa + o%. 
30. Multiply 4+ 5y by 2a + 3. 
Ans. 8a+ 10ay + 124+ 15y. 
31. Multiply 32? + 2y? by 227+ 3y’. 
Ans. 6 xt + 18 2°y’ + 6 y*. 
32. Multiply 2? + 3 a’y + 3827’ + y* by 2 —3 a’y + 3 2y’ 
—y’. Ans. x® —3 arty? + 8ax7y* — y’. 
33. Multiply 3 at —12 ae + 2az* —52* by Ta —2z. 
Ans. 2145 —90 ata + 24082? + 14 a07x> — 39 ax* + 102°. 
34. Multiply 3 a* + d5a’y + y’ by 3a* —5a’y + 7’. 
9a? —19 aty? + y*. 
35. Multiply 7 aa —2b’y + zx by 5a’ + 3 d* — y’. 
Ans. 35a'x —10a7b’y + 5a’xz + 21 a®bta —6 by + 
3 btxz —7T aay’ + 2 by? —ay’z. 
36. Multiply 16 atm? —da’e + 2? by a—z+ 327, 
Ans. 16 a®’m? —5a®x + az? —16a'm’z + 5arz —2 + 
48 atm?a? —15 ax? + 8 x27. 
37. Multiply 11 6’e — 5 a’y? + 22 by 14a3m + yz. 
Ans. 154 0°b’em —T0 amy’? + 28 a'mz + 11 Beyz 
— 5 ay*z + Qy2*. 
38. Multiply 13 7’s —5 rs? + 37s by 67's + 2 rs’ —rs. 
Ans. 78 rs? —47°s* + 5 73s? —10r’st -L 11 7’s* — 3 75%, 
39. Multiply 15 e’'d* — def + 2ed by 16 ef + 4ey —z. 
Ans. 240 c’dPef — 80 e? f? + 32 edef + 60 ctd®’y — 20c’efy 
+ 8e&dy —15 ed*z + 5 efz —2 edz. 
40. Multiply 16 atm + 2 am? —4 mi by 6 a’m — 83 am’, 
Ans. 96 a9m? + 12 a&m* — 24 a’m® — 48 aim? — 6 a?mi 
+ 12 am’. 


41, 


ELEMENTARY ALGEBRA. 57 


Multiply 25 a° + 6a'b + 7 a5b? by 2a —30. 


Ans. 50 at + 12a5b + 14 a%? —75 a*b —- 18 ath? — 21 ad’. 


42. 


43. 


44, 


45. 


46. 


Multiply 1 —3e@ + 32?—a2' by 1 —22 +4 a’. 
Ans. 1 —5x + 1027 —102° + 5a2t — 2’. 
Multiply 7a —3b by —5a + 26. 
Ans. —35 a’ + 29 ab — 60’. 
Multiply 5a? + 2ab —30’? by —5a-+ 68. 
Ans. — 25a’ + 2007b + 27 ab? —18 6%. 
Multiply 3 a? —4 ax + 52? by Ta? —2ax —38 a’. 
Ans. 21 at — 34 ax + 34072’? 4+ 2ax* — 152%. 
Multiply 5 a? —4ay + Ty’ by 2a? —3 ay —3y’. 
Ans. 10 at — 23 ay + 11la’y? —9 ay’ — 21 ¥'. 


. Multiply m* — n‘* by m* + nt. Ans. m® — n°. 
. Multiply m* —n‘* by mt —n*. Ans. m®’ —2 mnt + n°. 
. Multiply 82° —2.x2 + 5 by 4a*—a# + 6. 

Ans. 1227 —3a° + 182° —823 + 2227 —17x + 30. 
. Multiply 2 —ay + y —yz—az 4+ ? bye +y+2z. 


Ans. 22 + y§ + 22 —8 xyz. 


. Multiply 81 c+ 27¢y + 9c’? + 38cy> + y' by 8¢ —y. 


Ans. 248 —y’. 


. Multiply 32 a + 16a‘x + 8a'x’? + 407° + 2 axr* + 


x by 2a— x. Ans. 64a§ — 2°, 


. Multiply 1024 b> — 256 be + 64 dc? — 16d’ + 4 bet 


—éby4b+ 6. Ans. 4096 b§ — &. 


. Multiply 81c¢ + 54a + 36 ca? + 24 ca + 16 x* by 


3¢e—22. Ans. 243 & — 32 2°. 


. Multiply 75s° + rst + rs + 1 by —7°s* + 1’s’?’ — rs + 1. 


Ans. — 7°88 + 1. 


. Multiply m3 + 2 m’n + 2mn? + v3 by m? —2 mn + 


2 mn? — n3. Ans. m& — n§, 


. Multiply a7 — 2 a’y + 3a®y? —4aty® + 5a®yt —6a’y* 


+ Tay’ —8y' by & + 2ay 4+ y’. 
Ans. a —9ay® — 8 7’. 


. What is the value of (a? — ax + x”) (a? + ax -+ x”)? 


Ans. at + a?x? + xt, 


. What is the value of (« + 1) (1—2) (a + 8) (a —4)? 


Ans. st —2a° —13a° + 14x + 24. 


58 ELEMENTARY ALGEBRA. 


60. What is the value of (2 — 5) («a —6) (cw —7) (a —8)? 
Ans. x* —26 2° + 2512? — 1066 x + 1680. 
61. Expand (a + a) («7 + 5b) (x@ + ¢). 
Ans. 2° + ax’? + ba? + cx’? + abc + acx + dex + abe. 
62. Expand (x —a) (4 —n) (« —c). 
Ans. x° — ax* — na? —ca? + anx + acx + enxz —aen. 
63. Expand (a° —2*) (a + 2) (at —a’e + aa*? —az? 


+ 2*). Ans. 0" — a 
64. Expand (a + 2°) (a —«x) (at + ade + a2? + aa? 
+ a*), Ans. a” -—2 


65. Expand (a? — 2’) (a? + ax + x”) (a —ax + 2’). 
| ARS te 
66. Expand ab (a? + b?) (a+ 6) (a —b). Ans. a’b —ab’. 
67. Expand (a + b) (a —b) (a + 0’). Ans. 0 =e 
68. Multiply a —b + ¢ by a—b —e. 
Ans. 2 —2ab + B —e’. 
69. Multiply together x + y, y + z, and z+ w. 
Ans. vyz+ yz + 024 y2 + ayw + yw + xzw + yz. 
70. Multiply together 2 + y+ wandy + z+ w. 
Ans. cy + y? + Qyw + az + yz + zw + ow + w’ 


SECTION XV. 


Equations requiring Multiplication. 


ProsiEM 1. Given (x — 3) xX 10 = 40, to find the 
value of x. 


SoLuTIon. (x —3) x 10 = 40. 
Multiplying, 102 — 30 = 40. 
Transposing, 102 = 40 + 30. 

1032570: 


x= 7. 


ELEMENTARY ALGEBRA. 59 


ProsLEM 2. Given 5 + x x 11 + 14 = 91, to find 
the value of x. 


SOLUTION. 5+a¢x 11+ 14 = 91. 
Multiplying, 55+ lle+14= 91. 
Transposing, lla = 91 — 55 — 14. 

1M ra ps, 
eka 


ProBiEM 3. Find the value of x, when (6—z2z) (13 

—2x)—9 =a—8 X 2x4 35. 
SOLUTION. 
(5 —ax) (18 —22) —9 = « —8 X 2a + 385. 
Multiplying, 

65 — 23x" 4+ 227 —9 = 22’ —16a@ + 35. 

Here the 2x? may be cancelled in both members of the 

equation ; that is, subtracted from each of the equals. We 


then have 65 —232—9 = —162 + 35. 
Transposing, —23a2+ 162 = 35 —65 + 9. 
| —Tx# = — 21. 
If —7x = —21,then + 7x = + 21. This change 
may be made by multiplying the equals by —1. 
Hence, Tx = 21. 
rk Pes 


Norse. In problem 2d, 5 + zis multiplied by 11 only. If it were 
to be multiplied by 11 + 14, there would have been a parenthesis 
enclosing 11-+ 14. So, in problem 3d, z —8 is multiplied by 22 
only. 


EXAMPLES. 
Find the value of x in the following equations: 
1. (@ +6) x 9—2 = (@# —1) xX 40—1. Ans. ce = 3. 
2. (6 — ax) (6 —x) = (7 — x) (8 —x) — 10. 
| Ans. x = 4, 
3. (6 + x) (8 —7) = —9-a + 15. UY Rep ia 


ELEMENTARY ALGEBRA. 


. (@ —T) (w@ —8) = (@ — 2) (w@—9) + 1. 


Ans. = —2. 


(«+ 72+3=> (e—3)r¢—37. Ans. c= —4, 
.(@—4432)(5—2)=1l27—1. <Ansioeeah 





a—3 X 2£—20 = 2 —152 — 84. Ans. x = 18. 


7x + 7= 10a — 50. Ans. x = 19. 
YPt+aetli=wv’—b6er 8. Ans. x« = 1. 
. (5a + 3) (62+ 2) +6 = 2x24 9) (1542 — 2) + 30. 

Ans. x = 0. 
.2(3¢—10) = 54 —15. Ans. 2 = 5. 
.8x+7—(114+ 2) = 12. Ans. & = 8. 
. (« + 8) (x —T7) = (@ + 18) (a —12) + 50. 

Ans. x = 17. 
.e+ 11—2e27 = 52x —55. Ans. « = 11. 
-et+aer—20= 2’?—3a2 4+ 60. Ans. x = 20. 


(5a +1) (424+ 1) = (10e +1) Qe 4+ 1) — 18. 


Ans. x = 6. 


.3 = 62?—(22 + 2) (82+4+1)4+ 77. Ans. x = 9. 
. 34 —(2a + 20) = 6x4 — (xa + 80) — (8a + 40). 





Ans. x = 100. 
. («x —A) (4a —1) + 32 = (24 —2) (24 — 2) — 138. 

Ans. 0 =a 
. (7 + 2) 3 = («& — 2) 7. Ans. te cB. 
. (2 + 2) 4 = 9 (a — 2). Ans. x = 5}. 
.©@—6xX% 38=2—5x 17—4 . Ans. x« = 47%. 
-b9X38—a“2=7T—a2x 4. Ans. x = —13. 
. —13 (8 —32) = —11 (242). Ans. x = ti, 
.3xe—(x +1) = 7. Ans. ==, 
. 52 — (a — 38) = 30. Ans. © = Zo. 
.5—(@—1) = 6—CB-2 } 2). Ans. x = —1. 


.(8—62) xX —a#4+7 = —62? —(82 + 2) x — 4a. 


Ans. « = 74. 


(@+3)xHb+eoe—3% 64a) 


Ans. 2 = —3. 


, 2a + (2 —1) —(e + 1) —(#@—1) = 5. 


ELEMENTARY ALGEBRA. 61 


SECTION XVI. 


Questions producing Equations with One 
Unknown Quantity. 


ProBLEM 1. Two men together had $40. One gave the 
other $4 and then had 4 times as much as the other. How 
many had each at first ? 


Sotution. Let x stand for the number of dollars one 
had at first. 

Then 40 —~< stands for the number of dollars the other 
had at first. 

Then x —4 stands for what the first had after giving 
away $4. 

And 40 — x + 4 stands for what the other had after re- 
ceiving the $4. 

Now, by the conditions of the question, 
x—4 = 4 (40 —-a@ + 4). 


Multiplying, x—4= 160 —42 +4 16. 
Transposing, x+4xe=160+4 164 4. 
Collecting, 5a = 180. 

Dividing by 5, - == 30: 

Then, 40 —x = 4. 


Hence one had $36 and the other $4 at first. 


ProspLeM 2. Two men travel in opposite directions, 
starting from the same place, and at the same time. The 
first travels twice as many miles an hour as the second, and 
at the end of 5 hours they are 45 miles apart. How many 
miles an hour does each travel ? 


Sotution. Let x represent the number of miles an hour 
the second travels. 

Then 2 represents the number of miles an hour the first 
travels. 


Then 5 represents what the second travels in 5 hours. 
6 


62 ELEMENTARY ALGEBRA. 


And 102 represents what the first travels in 5 hours. 


Now, by the conditions of the question, — 
10x + 5a = 45. 


Collecting, 15 7 =="45. 
Taos 
Then, 2a = 6. 


Hence the first travels 6 miles an hour, and the second 3. 


Prose 3. A grocer sold 5 pounds of sugar, 4 pounds 
of coffee, and 28 pounds of flour, for $3.35. The price of a 
pound of coffee was twice that of a pound of sugar, and the 
price of a pound of sugar was 3 times that of a pound of 
flour. What was the price of each? 


Sotution. Let x represent the price of a pound of flour. 
Then 3 x represents the price of a pound of sugar. 
And 6 x represents the price of a pound of coffee. 
Also 28 x represents the price of 28 pounds of flour. 
And 15 represents the price of 5 pounds of sugar. 
And 24 x represents the price of 4 pounds of coffee. 
By the conditions of the question, 
282+ 152+ 24x = 335 cents. 


Collecting, 67 « = 335. 
c= 0, 

Then, 3 tee 

And 62 =a 


Hence the flour was 5 cents a pound; the sugar, 15; and 
the coffee, 30. 


<a 


ProBLEeM 4. One of two brothers is 4 years older than 
the other; and 6 times the age of the first is 8 years less 
than the product formed by multiplying a number 2 greater 
than the age of the second by 8. What is the age of each? 


Sotution. Let x stand for the age of the younger. 
Then # + 4 stands for the age of the older. 
Then 6 (w + 4) stands for 6 times the age of the older. 


ELEMENTARY ALGEBRA. 63 


And 8 (# + 2) stands for the last product alluded to in 
the question. 


Then, by the conditions of the question, 
6 (@ + 4) = 8 (@ + 2) — 8. 


Multiplying, 62+ 24= 82+ 16—8. 
Transposing, 62—8x = 16 — 8 — 24. 
Collecting, —2«“ = —16. 
Changing signs, pice Wi. 

x= 8. 
Then, a+4= 12. 


Hence the younger is 8 years old, and the older is 12. 


ProsLeM 5. A criminal escaped from prison, and tray- 
elled on foot for 4 days, when he stole a horse and travelled 
twice as fast as before. Five days after his escape, his route 
was discovered, and he was pursued at the rate of 70 miles 
a day, and overtaken in 4 days. At what rate did he travel 
on foot? 


SoLution. Let x be the number of miles he travelled on 
foot in a day. 
Then 2 x is the number of miles he travelled on horseback 
in a day. 
When he was overtaken, he had been out of prison 9 days, 
and had travelled 280 miles. 
Now 4 is the number of miles he travelled in the 4 days 
on foot. 
And 5 X 2« is the number of miles he travelled in the 
remaining 5, on horseback. 
Then, by the conditions of the question, 
4x2+ 102” = 280. 
Collecting, 14% = 280. 
==) 20: 
Hence he travelled 20 miles a day on foot. 


64 ELEMENTARY ALGEBRA. 


EXAMPLES. 


1. A boy bought the same number of tops, of marbles, 
and of balls, for 65 cents. For the marbles he gave 1 cent 
each; for the tops, 2 cents; and for the balls, 10 cents. 
How many of each did he buy? Ans. 5 of each. 


2. A grocer bought 5 boxes of raisins and 8 boxes of figs 
for $9. The raisins cost double as much per box as the figs. 
What did he pay for each? Ans. $1 a box for raisins, and 
50 cents a box for figs. 


3. Two men, 525 miles apart, begin at the same time to 
travel toward each other, one going 35, the other 40 miles 
aday. In what time will they meet? Ans. 7 days. 


4, What number multiplied by 5 gives a product as much 
greater than 30 as the number itself is less than 30? 
Ans. 10. 


5. A and B travel from the same place in thesame direction, 
A having 6 hours start. A goes 2 miles an hour, B 4 miles 
anhour. In what time will B overtake A? Ans. 6 hours. 


Note. Let x stand for the number of hours B travels to overtake A. 
Then z + 6 stands for the number of hours A travels. 

Now (x + 6) X 2 stands for the number of miles A goes. 

And « X 4 stands for the number of miles B goes. 


6. Out of 77 books, a certain number were sold, when there 
remained 3 more than were sold. How many were sold? 


Ans. 37 books. — 


7. Two men are a certain distance apart. They travel 
toward each other, one starting 3 hours before the other, 
and going 4 miles an hour, the other going 6. When they 
meet, it is found each has travelled the same distance. How 
long did each travel? Ans. The first travelled 9 hours ; 
and the second, 6 hours. 

8. Three pears, four lemons, and five oranges cost 43 
cents. A lemon cost 1 cent more than a pear, and an 
orange 2 cents more than a lemon. What was the cost of 


ELEMENTARY ALGEBRA.Y 65 
‘. ) 
each? Ans. A pear cost 2 cents; a lemon, 3 cents; and 
an orange, 5 cents. 


9. A, B, and C have $400 among them. A. has as much 
as B and C together, and B has 3 times as muth as C,and 
$4 besides. How much has each? Ans. A has $200, B 
has $151, and C has $49. 


10. What number exceeds its sixth part by 20? <Ans. 24. 


11. Divide $100 between A and B so that A shall have 
$20 more than B. Ans. A must have $60; and B, $40. 


12. Divide $600 among A, B, and CG, so that B shall have 
$30 more than A, and C $120 more than A. Ans. A must 
have $150; B, $180; and C, $270. 

13. Find two numbers whose sum shall be 60, and the 
first 8 times the second. Ans. 45 and 15. 


14. Find two numbers whose sum shall be 20, and their 
difference 8. Ans. 14 and 6. 


15. At an election the number of votes polled for two 
candidates was 1310. The successful candidate had a ma- 
jority of 262 votes. How many votes hadeach? Ans. One 
had 786 votes; the other, 524. 


16. Divide $4500 between A, B, and C, so that B shall 
have $1000 more than C, and A $1000 more than B. Ans. 
A must have $2500; B, $1500; and C, $500. 


17. In a certain company of 245 persons, there were 36 
more women than men, and 101 more children than both 
women and men. How many were there of each? Ans. 18 
men, 54 women, and 173 children. 


18. Divide an estate of $6000 among a widow, her two 
sons, and her four daughters, giving each son twice as much 
as a daughter, and the widow $2000 less than all her chil- 
dren together. Ans. The widow must have $2000; each 
son, $1000; and each daughter, $500. 


19. In a certain kind of military gunpowder, the charcoal 


and sulphur are each 4 of the nitre. How much of each 
6.* E 


66 ELEMENTARY ALGEBRA. 


ingredient is there in 800 lbs. Ans. 600 Ibs. of nitre, 100 
Ibs. of charcoal, and 100 Ibs. of sulphur. 


20. A man wished to enclose a piece of ground by a num- 
ber of palisades which he had. He found that if he set 
them a foot apart he would have too few by 40; but if he 
set them a yard apart he would have too many by 60. How 
many had he? Ans. 110 palisades. 

Note. If he set them a foot apart, he would require 3 times as 
many as if he set them a yard apart. Let x be the number he had; 


z+ 40 the number required for setting them a foot apart; x — 60 
the number required for setting them a yard apart. 


21. If a certain square piece of ground had been 5 feet 
longer and 5 feet wider, it would have contained 225 more 
square feet than it does. How many does it contain? Ans. 
400 square feet. 


Nott. Let z= its side; then x? — its area. If x+ 5 were its 
side, (x + 5) (x + 5) would be its area. 


22. If you add 16 to a certain number and multiply the 
sum by 3, the product will be 15 less than 6 times the num- 
ber. What is the number? Ans. 21. 

23. Divide the number 267 into two such parts that the 
greater diminished by 27 shall be 3.times the less. Ans. 207 
and 60. 


24. If B gives A $100, A will have 5 times as much as 
B. Both together have $360. How much has each. Ans. 
A has $200; and B, $160. 

25. What number is that from which 9 being subtracted, 
and the remainder multiplied by 13, the product will be 
117? Ans. 18. 

26. If the sum of a certain number and 9 be multiplied 
by 6, the result will be the same as if the sum of twice the 
number and 3 be multiplied by 8. What is the number? 
Ans. 3. 


27. Two persons, A and B, invest equal sums of money. 


ELEMENTARY ALGEBRA, 67 


A gains $700, and B loses $400, when A’s money is double 
B’s. What did each invest? Ans. $1500. 


28. A man goes to a tavern with a certain sum of money; 
spends $3; then borrows as much as he has left; again 
spends $3; then borrows twice as much as he has left; again 
spends $3 and has nothing left. How much had he at first? 
Ans. $5. 


29. A man goes to a tavern with a certain sum of money ; 
spends $3; then borrows as much as he had at first; again 
spends $3; then borrows twice as much as he had at first ; 
when he has $2 less than 3 times as much as he had at first. 
What had he at first? Ans. $4. 


30. A man goes to a tavern with a certain sum; spends 
$3; then borrows as much as he has left; again spends $3 ; 
then borrows as much as he has left; again spends $3; and 
again borrows as much as he has left ; when he has $6 more 
than twice as much as he had at first. What had he at first? 
Ans. $8. 


SECTION XVII. 
Division. The Signs + and —. 


Drviston is the process of finding how often one quantity, 
called the divisor, is contained in another, called the dividend. 
The result is called the quotient. 

It is the reverse of multiplication. The divisor multiplied 
by the quotient must always produce the dividend. 

If we divide 12 by 4, the quotient is 8. 12 + 4 = 3. 
We may divide + 12 by + 4 or by —4; we may also di- 
vide —12 by + 4 or—4. Remembering that the divisor 
multiplied by the quotient produces the dividend, we may 
find what sign the quotient must have in all the cases. 


68 ELEMENTARY ALGEBRA, 


divisor 
Divide + 12 by + 4. The quotient is + 3, because + 4 
quotient dividend 
multiplied by + 3 = + 12. 


divisor 
Divide + 12 by —4. The quotient is — 3, because — 4 
quotient dividend 
multiplied by —3 = + 12. 
divisor 
Divide —12 by + 4. The quotient is — 3, because + 4 
quotient dividend 
multiplied by — 3 = —12. 
divisor 
Divide —12 by —4. The quotient is + 3, because — 4 
quotient dividend 
multiplied by + 3 = —12. 


Here we see that when the dividend and divisor have like 
signs, the sign of the quotient is +; and when they have 
unlike signs, it is —- 

The same reasoning wea apply to any operation of division. 
Hence, whenever we divide one quantity by another, like 
signs give +, and unlike signs give —. 

Note. Examples in Division may be proved by multiplying the 
divisor by the quotient. The result should be the same as the divi- 
dend. 


EXAMPLES. 
1. Divide + 9 by + 3. Ans. 3. 
2. Divide + 9 by —3. Ans. — 3. 
3. Divide —9 by —3. Ans. 3. 
4, Divide —9 by + 38. Ans. — 3. 
5. Divide + 27 by — 9. Ans. — 3. 
6. Divide — 27 by + 9. Ans. — 3. 
7. Divide —6 x by 2. Ans. — 32. 
8. Divide —6x by —2. Ans. 3x. 
9. Divide —4a by 2a. Ans. —2. 
10, Divide — 68 y by 9. Ans. —Ty. 


ELEMENTARY ALGEBRA, 69 


11. Divide 63 y by 9. Ans. 7. 
12. Divide 63 y by —9y. Ans. —T. 
13. Divide — 63 y by —9 y. Ans. 7. 
14. Divide 3 ae by ac. Ans. 3. 
15. Divide — 25a by — 25. Ans, xX. 
16. Divide — 42c by 42. Ans. —ce. 
17. Divide 12 xy by — zy. Ans. —12. 
18. Divide —14 ab by —ab. Ans. 14. 
19. Divide 22 a by — 22. Ans. — a. 
20. Divide —72 by 12. Ans, —6. 
——0t@400-——. 


SECTION XVIII. 


Division. Letters without Indices. 


Ir must be remembered that letters stand for numbers, 
and that writing them together indicates multiplication. 
Thus ae is the product of a andec. If a stands for 5 and ¢ 
for 4, ac must stand for 20. 

If we divide the product ac by a, the quotient ise. Ifwe 
divide ac by c¢, the quotient is a. 

So, vy -~ x= y; andsy +y=-. 

Also, npx + & = np; npx + px =n; and npx ~ nx 
= p. 

Thus we see that to omit one of the factors of a term is to 
divide by that factor. 


EXAMPLES. 
1. Divide abe by ab. Ans. ¢. 
2. Divide 8 abx by ab. Ans. 8 x. 
3. Divide 8 abx by 4ab. Ang 2a. 
4. Divide —8 abx by —ab. Ans. 8 x. 
5. Divide ed by ed. Ans. 1. 
6 


. Divide —cd by —cd. Ans. 1. 


7. Divide —ed by + cd. Ans. —1. 
8. Divide aa by a. Ans. a. 
9. Divide 4 aa by a. Ans. 4a. 
10. Divide 4 aaa by —a. Ans. — 4 aa. 
11. Divide —7 aaaaa by — aaa. Ans. 7 aa. 
12. Divide 24 aaaxx by — 6 aaz. Ans. —4an. 


ELEMENTARY ALGEBRA, 


. Divide — 12 ecceecy by 4 ce. 
. Divide — 50 rxzrrrrx by 5 xrxx. 


Ans. — 8 ceecy. 
Ans. —10 xxx. 


15. Divide — 10 abyyyyy by —5 abyyy. Ans. 2yy. 
16. Divide 17 abbbyy by —17 bbyy. Ans. —ab. 
17. Divide —9 aaxxx by 3 aax. Ans. —3 xx, 
18. Divide 20 aaacce by 5 aaacce. Ans. 4. 
19. Divide 14 abry by — 14 abay. Ans. —1. 
20. Divide —10cn by —1. Ans. 10 cn. 


SECTION XIX. 


Division. 


Monomials with Indices. 


We have seen that xarxxvnx ~— xnxx = vax. The same 


thing may be expressed thus, «' + 2‘ = a*. The exponents 
or indices, 7, 4, and 3, only show how many times z is taken 
as a factor. And, as division is the reverse of multiplica- 
tion, it must be as many times a factor in the quotient, as 
the number in the dividend less that in the divisor. Hence, 
‘with powers of the same letter, subtract the exponent in the di- 
visor from that in the dividend. The remainder will be the 
exponent of the same letter in the quotient. 

It is necessary here to observe that any power which has 
0 for its exponent is equal to 1. This is shown by dividing 
any power by itself. For instance, divide x’ by x’. Sub- 
tract the index of the divisor from that of the dividend, and ~ 
we get 0. Hence, 2? + a’ = 2° But we know that any- 


ELEMENTARY ALGEBRA. 71 


thing will go into itself once. Hence, 2’ + x7=1. There- 
fore, x® must be equal tol. In the same way, a? = 1, n° 
= 1, any power which has 0 as its exponent is equal to 1. 
So, when we divide ab‘ce by abtc’, we may set down the quo- 
tient a°b°*, which is the same as 1 X 1 X ©, or simply ¢: 
for when 1 is a factor it need not be expressed. So, again, 
24 a®bic’d*e® — 4a%bte’'d* = 6 abdte’. 


EXAMPLES. 
1. Divide 2° by x*. Ans. x*. 
2. Divide 9 x* by —3 x, Ans. — 8 x. 
3. Divide —x* by z*. Ans. — 2°, 
4, Divide — a* by — 2%. Anas &, 
5. Divide — 2 by x*. Ans. — x*. 
6. Divide 7 x by 2’. Ans. 7 2°, 
7. Divide 6 a’x* by — a’x’. Ans. — 6 ax”. 
8. Divide — a®x*y by — aby. Ans. x. 
9. Divide 11 atzy’? by — a’. Ans. — 11 a?xy’. 
10. Divide 15 27y°z? by 5 xz. Ans. 3 y*z. 
11. Divide — a’c'd" by — a’c*d?. Ans. wedt. 
12. Divide 24 a®c’d® by 8 acd’. Ans. 3 a®e. 
13. Divide 15 a?a*y* by — 5 ax’*y®. Ans. —3axy. 
14. Divide — 10 a®x?y’ by — 2 a®x*y. Ans. 5 ay. 
15. Divide — 14 m'n*p*¢’ by 7 m’np*. Ans. — 2 mn’. 
16. Divide —10 a y°2° by —5 ayz®. Ans. 2 at. 
17. Divide 12 a*bctd? by — 4 a®b'd?. Ans. —3¢. 
18. Divide — 16 a®b‘c’d by 16 a®c'd. Ans. — bte, 
19. Divide 19 a®b‘x’y by — 19 abx’*. Ans. —a'b’y, 
20. Divide — 4 x*y°z? by — 2’y*. Ans. 4 xy2’.. 
21. Divide — 16 p°q’r's°t by p'qr. Ans. — 16 qr’s't. 
22. Divide 9 a®p*t?a* by 9 apt’. Ans. apa’. 
23. Divide — 21 A’k’m‘n® by — 7 h®k?m*nk, Ans, 3. 
24. Divide 32 a®bq*r’ by — 8 atbr. Ans. —4 a‘g’r. 
25. Divide 15 m’ntxty? by 5 min’y’. Ans. 3 m’nz-+, 
26. Divide —12 a®b‘c'd’? by — 4 a*bd®. Ans. 3 abe. 


27. Divide — 25 p*q*r*s*tu’ by 5 p'gtr®s’. Ans. —5 st us 


72 


98. 
99. 
30. 
31. 
32. 
33. 
34, 
35. 
36. 
BY: 
38. 
39. 
AO, 


Divide — 24 weuty? by — 8 uw’. 
Divide 89 f*g*h®k by 18 f*hk. 
Divide 49 hkp*q* by —7 hp’q’. 
Divide — 638 x’yz’u by 7 xyzu. 
Divide — 21 x*ab‘e by 3 x’ab. 
Divide 93 m’ya®xb by — 31 m’az. 
Divide 81 ay’x*e? by 9 yarae. 
Divide — 18 zxtay’d by — 2 xday’. 
Divide — 24 frga by — 38 fa. 
Divide — 12 f*g*a°a by 12 f%g*aa’. 
Divide 16 m’npg by —4 m’np. 
Divide 12 a*xc’b by 3 axbe. 
Divide — 33 &ax*m by 11 aca’. 


ELEMENTARY ALGEBRA. 


Ans. 3 uty’. 

— Ans. 3 ght. 
Ans. —7 kpgq. 
Ans. —9 xz. 
Ans. —7 b’ca. 
Ans. — 3 aby. 
Ans. 9 aca'y. 
Ans. 9 x*y. 
Ans. 8 ag. 
Ans. —1. 
Ans. — 4 mq. 
Ans. 4’. 
Ans, — 3m. 


When any index in the divisor is greater than the corre- 
sponding one in the dividend, it is best to set the divisor 
under the dividend in the form of a fraction, and cancel all 
common factors, that is, divide as far as can be done. To 


divide a’ct by a®c, we set down 


a’ct, 


=o an 
ac 


d cancel @ and ¢e in 


8 
both numerator and denominator. The result is be 
a 


41. 


42. 


43. 


44, 


45. 


46. 


Divide 12 a°b3x* by 6 a‘b?x. 
Divide 13 a’y*z? by — 25 ayz’. 
Divide — 9 x’y°zt by 4 xyz. 
Divide 14 m’n’x> by — 3 m‘nx®. 
Divide —7 a®b'c’d'e? by 35 ab¥ct, 


Divide — 93 abate’d by — 36 xd. 











Ans. oie 

Ans. — = ; 
Ans. — ix 

Ans. = ia se i 
3mx 

Ans. — Ss 
Ase 31 abc? 


47, 


48. 


49, 


50. 


51. 


52. 


08. 


‘4, 


55. 


56. 


57. 


58. 


59. 


ELEMENTARY ALGEBRA. 


Divide 21 m‘n*q’a* by 35 ma’. 
Divide 33 a*y’m?at by — 638 am*a?. 
Divide — 27 x°m’y* by 36 m’yz*. 
Divide 13 abcde by — 25 eax. 
Divide 14 atsx*z?m? by — 35 a®z’q. 
Divide — 15 f%9°q"* by —5 fig’. 
Divide — 21 aty’z'w by 65 z°yw. 
Divide 19 w’z*v by 23 ubw’. 
Divide 12 x°c°b?y* by a 34 c°b*. 


Divide 24 abte? by 13 d. 


Divide —18 m’n’x‘p by — 21 m‘z%p. 


Divide — 19 y’x by —19 y’px. 


Divide —15 ab by 50 6’. 
7 


Ans. — 





Ans. on 


11 an*y" 
OT eae 


xy® 
4 ~ 








Ans. — 








24 a®bte? 
8. i3d . 





6 nba 
Tm ° 


Ans. A 
p 


Ans. 


oo 


Ans. — 0s 


74 


60. 


61. 


62. 


63. 


64. 


65. 


66. 


67. 


68. 


69. 


70. 


ELEMENTARY ALGEBRA. 


Divide 18 abed by — 21 ed’. 
Divide — 36 7’s*tt by — 45 rés®x. 
Divide 34 p’qrs’? by 51 p’q’tz’*. 
Divide — 48 a*y’z* by 135 m®a’y*q. 
Divide —12 v’wutx by — 15 wa. 
Divide 49 a®y?m'n* by — 7 x?min’. 
Divide — 60 m“n*p” by 125 mp”. 
Divide 25 gh” by 35 hk, 
Divide — 13 w’y*z by — 8 a’x’y'z. 


Divide 15 c*d’e*f by 25 g®e?fe*. 


Divide — 33 x’y%z*m by — 33 a’y*z*m. 





6 ab 


AN eae 








ELEMENTARY ALGEBRA, 75 


SECTION XX. 


Division of a Polynomial by a Monomial. 


WHEN the dividend contains several terms and the di- 
visor only one, each term of the dividend must be divided. 


Prosiem. Divide 34a*zx'y —12 axy + 3ay' by 2 a’xy. 


OPERATION. 2a’xy ) 34 aa'y —12a@7xy + 32y'* 





: 3y° 
17 ax* — 6 + , 
Pa iy 
EXAMPLES. 
. Divide ax + bx + ex by x. Anjwatb+e. 


2. Divide —9 aty + 6 a'y —3a’y by —3 xy. 


10. 


Ans. 32° —22? + x. 


. Divide 20 ac? — 45 aca + 10 a'xy by 5 a’. 


Ans. 4c? —9cex + Qaaxy. 


. Divide 15 aby —9 acy’ + 12 axy® by 8 a’y. 


Ans. 5 ab —8 cy? + 4a*y'. 


. Divide 36 ab’x*y — 16 2°y? — 20 x*y* by — 427y. 


Ans. —9a’b? + 4a'y + 5 xy’. 


. Divide 4 a’z* + 7 art — 9 atxd by — az’, 


Ans. — 4axn —7 a®a? + 9 a323. 


. Divide 6 atz*m? —8a*bx?m? + 10 aSerm? —4 abdx?m? 


by 2 a®a5m?. Ans. 8a —4) -- 6c —2d. 


. Divide — 15 ata*yt — 9 ata*y’ + 6 a®aty? by — 3 ataty’. 


Ans. 5y? + 32° —2 at, 


. Divide 75 ac'x? + 25 a2ce®'a —15 ac’a*yz by 5 eax. 


Ans. 15 acx + 5 ac —8 xyz, 
Divide 6 atem — 9 atem? + 12 atem® by — 8 a’, 
Ans. —2am + 3 am? — 4 ami. 


76 


11. 
12. 
13. 
14. 
15. 


16. 


17. 


18. 


19. 


20. 


21. 


22. 


ELEMENTARY ALGEBRA, 


Divide 7 ackx + 7 ac’y —7 aéz —T aed by 7 ac’. 
Ans. x + y—z —d. 
Divide 45 am‘n? — 27 am’n* + 63 m'n'y by 9 min’. 
Ans. 5am —8an + 7 n’y. 
Divide 36 p’q’rs’ — 72 p’q’rt? + 84 pq’ra® by — 12 pq’. 
Ans. —3 prs? + 6 pr? —T ra’. 
Divide 42 am?r*st? —63 be’r*st? + 105 7°s*?x by —21 7st”. 
Ans. — 2am + 3b — 58a. 
Divide 126 atx’y — 30 a®b’x’y + 18 ax*y’ by 6 xy. 
Ans. 21 at —5 ab? + 3 ary. 
Divide 49 vw’st — 35 vw®.— 2 2’yw* by 7 va’. 
Su Quty 
as 7 vx 


Divide 63 e?f*y + 27 etf*a'z — 7 éfax* by 9 efx’. 





Ans. 7 wx — 


Tee 


Ans. es J 48 ef xez — 
Divide 32 pq*rs — 32 cp’r — 48 p*q’r by — 16 pq’r. 
Ans. —2q's + e + 3p’ 
Divide —18 a®y + 16 y® + 27 a®y°z by —3 ay. 
Ans. 6a — eke — 9 a’y*z. 
Divide — abex + abe’x? —be by — abex. 
Ans. 1 —cx + = 
Divide 5 a’c'x? —7 aa*y + 3 2’cd by a’c’x*, 
Ans. 5¢ — ee) 2 


ac 
Divide — 21 foh + Tf?g —5 fg’ by —38fogh. 
Ans. 7 — 








23. 


24, 


20. 


26. 


27. 


28. 


29. 


30. 


ELEMENTARY ALGEBRA. 77 


Divide 4 y’cx —2 pat —y’ by 2 pe. 


9 2 
Mngt a po — woe ; 
Hs 








2 px 
Divide 20 a®atz + 16 a’x’z? — 3 mnb by 5a*x*z. 
16 z 3 bmn 
5 2 rae Se ko 
Ans. 4a°a? + 5a? 5 gtxte’ 
Divide 36 fa’e —5r’s’x + 63 fex*c? by 21 f%x’e. 
12 ores: og 
Ans. 7fi— Weft + 3 f*x’e. 
Divide 36 am’z* — 54 2’a + 34atm by —18 zam. 
3 17 a* 
Ans. — 2 mz ote ae =. 9 az 


Divide 16 cfx* + 20 cfa®y —3 fretrtz by 4 efx’. 
Ans. 4c” + 5exy — pa 


Divide 18 amz*> + 9m — 15 a’mix’y by 5 a’mtx. 


1 Ea yfirs h , 
Ans. Bat + —— —s8amey. 


5a’mx 


Divide 32 abede —5 axed + 3e by —2 abcde. 


5 ax 3 8 
Ans. —16 A 5 ta a 
ae a 2 be 2 abed 


Divide 25 a®d’e’x + 21 ated — 30 atdex® by 5 abdex. 


Ans. 5a’de + ae — 6 ax’, 





7% 


78 


ELEMENTARY ALGEBRA, 


SECTION XXI. 


Review of Subtraction. 


BEFORE proceeding any further in division, the student 
should make sure that he has fully mastered subtraction. 
The following examples are intended to afford exercise in 
such difficulties of subtraction as are likely to be met with 
in division. 


“Im or > 


. From a + 2azx take a? + ax. Ans. ax. 
. From @ + 2ax + a’ takea? + az. Ans. ax + 2%, 
. From w@ —2 ax + 2’ take a + ax. 


Ans. —3ax + a”. 


. From a? —2axz + x’ take a? —ax. Ans. —ax 4+ a”. 
. From ax + a’ take ax 4+ 2’, Ans. 0. 
. From —az + 2’ take —ax + 22. Ans. 0. 


. From 6 a7a? + 4 ac° + at take 6 a2? + 3 az’. 


Ans. ax + at, 


. From — 8 aa —5a’*x? + 38a2a° —2* take —8 a’a + 


38 a7x2 —38ax°+ 5at, Ans. —8a’7x + 6ax® —6 2+. 


. From —6a’°y — 3ay —2y*> take —6x’*y — 227’ 


—4y’. Ans. 8 ee 


. From —5 arty + 7° take —5 aty + 10 ay’ 


Ans. — 10 xy? x cot 


: teh —4Aaa + 6a*x? —4a2z° + a* take at — 2 a®x 


+ a’x”, Ans. —2a?x + 5 a®x? —4aa° + ot, 


. From —2 ax + 5 ax? —4a2° take —2a*x + 4072’ 


— 2 ax’. Ans. a?z? — 2 az. 


. From a’x? —2ax ake a?x? — : 
F "7 —2Qax*> + a take ax? —2ax* + x 


Ans. 0. 


. From —2’y? + y’ take —a’yi—ay*. Ans. zy* + y’. 
15. 
. From —4az + 62’ take —4 ax —9 2” + y. 


From ryt + y° take ryt + y’. Ans. 0. 


Ans. 15 x? —y. 


ELEMENTARY ALGEBRA, 79 


17. From — 40 a°b + 57 «2b? + 4ab' —15 b¢ take — 40 a%b 
+ 320°)? + 24ab% Ans. 25 ab? —20.ab® —15 b+ 
18. From —2a°b —3a’b? —6ab®? + 40 take —2a°d 


—-4q'>? —. 8 ab’. Ans. a7b? + 2ab*? + 454, 
19. From —5a@’8? + 16 ab? —4 0‘ take — 5 a’?h? —4 0+. 
Ans. 16 ab’. 


20. From 6 ab? — 5 abt + 4ab' take 6 ab? — 4 ab®, 
Ans. — 5abt + 8ab>. 


——.059300—— 


SECTION XXII. 
Division of a Polynomial by a Polynomial. 


WHEN both dividend and divisor have several terms, the 
process is much like long division in Arithmetic. 


ProBLEM 1. Divide cx + bx by c + Bb. 


OPERATION. 
divisor dividend quotient 
e+tb)cx+ bx(z 
cx + bx 


EXPLANATION. The first term, c, of the divisor, is con- 
tained in the first term, cx, of the dividend, x times. Mul- 
tiplying the whole divisor by 2, the product is found to be 
ex + bx, which we subtract from the dividend, and find no 
remainder. 


ProsuLeM 2. Divide mz + my —cx —cy by x + y. 


OPERATION. 
x+y)mxe + my —cxr —cy (m—e 
mx + my 
— cx —cy 
— cx —cy 


EXPLANATION. Here x goes m times into mx; hence we 


80 ELEMENTARY ALGEBRA. 


set down m in the quotient. We then multiply the whole 
divisor by m, and subtract the product from the dividend. 
The remainder is —ca —cy. This also must be divided by 
x+y. Wesee that x goes —c times into —cx; hence we 
set —c also in the quotient. Proceeding as before, we find 
no remainder. 


ProBLEM 3. Divide —3a@’x —a* + 3a2? + a by a 
+ x2? —2ax. 
OPERATION. 
a —2Jax + x’) a —3ax + 3ax? —2° (a—ex 
a —Qax + ax? 
—avx + Qax? — 2? 
—ax + 2ax? — x’ 


EXPLANATION. In this problem, the only letters are a 
and x raised to various powers. One of these letters should 
be selected, and both dividend and divisor arranged so that 
the highest power of that letter comes first, then the next in 
order, and soon. This is for convenience. In the forego- 
ing operation, the terms are arranged according to the powers 
of a. In the divisor, we begin with the second power of a, 
next comes the first power, and finally a term without a. In 
the dividend, we have the third power, the second power, 
and the first power of a, and also a term without a. The 
process of dividing is the same as before. 

Let us now perform the same example, arranging the 
terms according to the powers of x. 


OPERATION. 
ow’ —2Qarn +a )—2+ 38ax*°—8ar+a(—x+a 
Lidar ae 


ax? —Qa’x + a’ 
ax’? —2ax + a 


Nore. It will be observed that the quotients, a —z and —z + a, 
are precisely the same. 


ELEMENTARY ALGEBRA, 81 


ProsuiEM 4. Divide «* —w* by « —w. 


OPERATION. 
x—w) x —w? (at + a?w + aw? + aw? + wt 
ao’ — aw 
xtw —w 
xstw —x*w? 
rw — up 
ew? — ow? 
x’w° —w? 


x’w®? — rw 
xcw*t —w 


awt —w? 


EXPLANATION. In the dividend, the fourth, third, sec- 
ond, and first powers of « are wanting. Space has been left 
for them in the operation. 


ProsueM 5. Divide x*? + ybyx+y. 


OPERATION. 
a a +y(e+eryty 
x* — xy 
wy + y 
xy axe xy’ 
sy + y° 
cy Ye 








2 y* remainder. 


EXPLANATION. The quotient may be completed by a 
fraction whose numerator is the remainder and whose de- 
nominator is the divisor. The full quotient is x* + ay 

2y’ 
eed 


aie 4 


82 ELEMENTARY ALGEBRA. 


ProsLEM 6. Divide 1 by 1 + 2. 








OPERATION. 
eee yt (1—2 + ata? + at, 
1+2 
—wzx 
—r—xz 
a? 
ot + x 
en SP 
— x — xt 
at 
at + a 





— 7 


EXPLANATION. This operation may be carried as far as 
we please in the same way. When we have carried it as 
far as desirable, the quotient may be completed as in Prob- 


lem 5th. Observe that its sign is — in the above operation. 
5 





; : A F as 
The full quotient, then, is 1 —a# + 2? —a* + a* — eras 
EXAMPLES. 
1. Divide @ + 2ax+ a2’? byat oa. Ans. a + 2. 
2. Divide a? — 2 ax + 2? by a — «. Ans. @ —®&. 
3. Divide a —2’ by a + x. Ans. & —%. 
4. Divide m? — n? by m —n. Ans. m+ n. 
5. Divide ae — be — ad + bd by a — 6. Ans. ce —d. 
6. Divide —7r’? + s’ by —r+ s. Ans. r+ s. 
7. Divide at + 2a’x? + ax by a 4+ a”. Ans. a? + 2. 
8. Divide a* —y* by x —y. Ans. x + xy + y’. 
9. Divide zt —y* by x —y. Ans. 2° + a’y 4+ ay? + y’. 
10. Divide a’ — 0° by a — 2b. 
Ans. a + ab + a®d? + a5? + abt + BE 
11. Divide a’ — 3 a‘b? + 8 abt — B by a& — 30°) + 3ad’ 


— 6, Ans. @& +307b + 38ab? + BF. 


12. 
13. 
14, 
15. 
16. 
AVE 
18. 
19. 
20. 
21. 


22. 


ELEMENTARY ALGEBRA. 83 


Divide m’ + m’n + mn? + n3 by m + nr. 
Ans. m’ ok rn’, 
Divide m* + m’n + mn® + nt by m + n. 
Ans. m® + ni’. 
Divide a + 1 bya +1. Ans. @ —a-+1. 
Divide y® — y by y’ — y+ y® — y. Ans. y* + 1. 
Divide 6° —1 by 6? —1. Ans. 6§ + 66+ 6? + 1. 
Divide 36 a’ — 36 B' by 9a — 908. 
Ans. 4a® + 40°) + 4 ath? + 405? + 40°b* + 48° 
+ 4 b% 
Divide 2* — 256 by 2? —4. 
Ans. #6 + 4a* + 162’ + 64, 
Divide 16 «* — 81 y* by 24 —3 y. 
Ans. 8a + 12 a’y + 18 ay? + 27’. 
Divide 16a*+ 8a-+ 38 by 4a? —4a-+ 3. 
Ans. 47 +4a+1. 
Divide 81 2* — 81 x2? — 36 ry —4y’ by 92?+ 9u + 
2 y. Ans. 92° —9x —2y. 
Divide 64 c&& — 128 ed*® + 64 d* by 2¢ —2d. 


Ans, 82+ 32ctd + 32 ed? —32 cd’ — 32 edt — 32 d°. 


23. 
24. 


25. 
26. 


27. 


28. 


29. 


Divide 18 2’y? —15 xy + 2 by 6 xy —1. 





Ans. 8 ae — 2. 
y? 
Divide 2? 2b A — . 
+ y yret+y Ns. X dare 
Divide 2° + y°> by x + y. Ans, 2 — Ty -- ¥ 
Divide «* + y* by x + y. 
2 v4 





Ans, 2 —2*y + ay?—y + 
Divide x® + a by x + a. 


Ans. at — ax? a5 ax? — va + at 


Divide a® + b& by a + 0. 
Ans. a —a‘b + a®b? —a’b® + abt —b® + —_. 


=; b 
Divide c’ + d’ by ¢ + d. 
Ans. & —ed + cfd? — ed + ed‘ —ed® + d’ 


o+ty 


84 


30. 


31. 


32. 


33. 


34. 


30. 


36. 


37. 


41. 


42. 


ELEMENTARY ALGEBRA, 





Divide m$ + n° by m + n. Ans. m' —m'n + mn} 
— mn? + mnt — mn? + mné 2 n* 

m + n 

Divide 96 xy? + 12.a°y* — 24 aby’ — 48 ay’ —6 ays 


+ 12 ay" by 6 ay —3 zy’. 

-Ans. 16 aty + 2ry? —4 yy. 
Divide 32 — 80x + 8027 — 40 a#* + 102+ —=25 by 4 
—4a+ x’, Ans. 8 —12a + 62? — a". 
Divide — a®b® + 1 by —a*b* + ab? —ab + 1. 

Ans. a®b® + abt + ab + 1. 

Divide 312 xy? — 32 zy’ + 20 wy? — 160 2’y* + 88 
xy’ — 12 0°y? by 12 a°y + 8 ry’ —2 ay. 

Ans. she — 20 xi" a 6 xy. 
Divide x? —9 ay® — 8 y? by ay? + 2ary’? + y’. 
Ans. x'y’ —2 ay’ + 3 aby’ —4 vty’ + 5 ay’ —6 xy" 





+ Tay’ —8y'. 
Divide 243 m'n§ — mn” by 81 mint + 27 min? + 9 
mon® + 8mm" + mén’, Ans. 3mn— mn’. 
Divide a? —b*® by a + 0. an 
anaes 2 SS ee 
Ans. a —ab + b ai ; 
| ig 
ae Pale ee ae 
. Divide 1 —22’> by 1 +2. Ans. 1 —x Tae 
y? 
2 ‘ 2 
. Divide x? by x + y. Ans. x vt ae 
. Divide 1 + 4a by 1— 6-2. 
Ans. 1+ l0e+ > 60 x nb 
—62 
Divide x* + 32° by 2? —2a 4 1. 
9a” —5 
Ans. x + 5 ae 3D eee 
Divide 182 — 44a? — 24a + 3a° + B by 6— 2a’. 
E Ans. 22—4a + : oe 


6—2a? 


43. 


44. 
45, 


46. 


47. 


48. 
49. 
50. 
51. 
52. 
53. 


54. 


5d. 
56. 


o7. 


ELEMENTARY ALGEBRA, 85 


Divide 56 y? — 4 by — Bb’ by Ty — b. 








4 by —B’ 
Ans. 8y + a ear 
Divide 162° — 962’? + 40% —240+4+ 2° + y' by 22 
oa 19, Ans. 80° + 20 + et 
Divide #* + y°+ zbyx+y. 
2 y'* tie 
Ans, or zy”? — 
n y + ry’? —y + aT; 


Divide 20 a7b — 25 a —18b* + 27ab? by —5a + 66. 
Ans. 507 + 2ab —3 & 
Divide 382’y? + 48 cid’x’? 4- 5eay + cy? + 16 ed? 

—dceur —16 cd’y —y’ — 15 ea’ by c—y4+ 82. 
Ans. 16 cd? — 5e’x + y*. 

Divide — 9 a’y? + 25 a’y’? by 5a®y + 38 aty. 

Ans. 5 ay —3 aty. 
Divide —3 ed® + 4ed —9c’d? + 6ct+ 2d‘ by — ed 


+ 3¢ —2 ad’. Ans. 2c? + 2cd — d?. 
Divide 12 x’2? + 5y4 —32 2°’ + 48 ct —3 y?2? by 42° 
—y’. en 122°— by 4+ 327, 
Dividea + b by a+ 8. Ans. 1 


Divide ay + by + ax + ba bya+6. Ans. yt oa. 
Divide 6 cz? —16 ca’'y — 56 c’aty’? + 18 «y* by 2 cx 
— 8 cu’y — 6 xy’. Ans. 80’ + 4 cx’y — 3 xy’, 
Divide 28 —2a7 + 32° —3a° + 32* —32° + 32? 
—2¢+1 by #’—2+4+1. 
Ans. c& —a> + ot —a? + a’? —o + 1. 
Divide 1 by 1 —a + 2’. 
. Ans. 1 + «@ —2? — ot 4+ of + a’ — Ke. 
Divide 1 + 6z by 1—2z. 


Ans. 1 + 8x + 162? + 322° 4+ 642+ + 12825 + &e. 


Divide 1 by 1 —4z. 


Ans. 14+4a24+162?4+ 642° + 256 a + 1024284 &e. 


58. 


Divide 1 + 7x by 1+ 6z. 
Ans. 1 + «1 —6 2? + 36 2° — 216 xt + 1296 a — &e. 


8 


86 


59. 


60. 


61. 


62. 


63. 


64. 


60. 


66. 


67. 


68. 
69. 


70. 


12 


72. 
73. 


74. 
70. 


ELEMENTARY ALGEBRA. 


Divide 82* —8a’y? + 38272 + byt —38y'2 by 2 


—y’. Ans. 32? —5y’ + 32’, 
Divide 4 «* — 25 ay* + 20 ry’ —4 y® by 2 2° + 5 xy’ 
— 27’. — Ans. 243 —d ay? + 2y’. 
Divide 6 x#’y — 12 ryz + 6y2 by 2xy —2 yz. 
Ans. 8% —3z. 
Divide m? — mx — x —1 by m —a —1. 
Ans. m + 1. 
Divide 4p* — 25 p’¢* + 20 pq’ —4¢q° by 2p? — 5 pq’ 
+ 2 q’. Ans. 2p? + dpq’ —2 4. 
Divide ax + cx + ay + bx + bz + by +z + cy t+ 
azbyx+y+z. Ans atb+e. 


Divide ath? — 2 a®b®? + a’b* by a’*b — ab’. 
Ans. a’b — ab’. 
Divide ab? + ab? + a’b + ab’ by ab? + ab. 
Ans. a + b. 
Divide — 16 a° + 36 x*y’ by 4a? + 627y. 
Ans. —4a® + 6 x’y. 


Divide a® — a’ by a? —1. Ans. a’ + a’, 
Divide 6x* + 4a°y —9 ax’y’ —3 ry? + 2 y* by 2x? + 
2xy —y’. Ans. 3 x —ay —2y’. 
Divide « + 32a) by x + 2a. 


Ans. xt —2a°a + 4x70? —8 xa? + 16a. 


Divide m + n by a? — mn. Ans. Baers 

a> —mn 

Divide acy —pey by a —p. Ans, cy. 

Dividea+ aby a+ a. Ans. 1. 
2 

Divide n? + x’ by n? — 2. Ans. 1 + pee 


Divide a® — b* by a” —b. 
Ans. a® + ab + ab? + 6%. 


Note. Examples in Division may be proved by multiplying the 
divisor by the quotient. The result should be the same as the divi- 
dend, 


ELEMENTARY ALGEBRA, 87 


SECTION XXIII. 
Equations with Two Unknown Quantities. 


PROBLEM 1. There are two numbers whose sum is 43 
and whose difference is 11. Find them. 


Sotution. Let x stand for the greater number, and y for 
the less. 


Then, according to the conditions of the problem, 








x+y = 43. 
And L—y= Ui 
Adding equals to equals, 2 x = 54, 
Dividing by 2, x aie 
Putting 27 instead of x in the first equation, we have 
27 + y = 48. 
Transposing, y = 48 — 27. 

y = 16. 


Hence the numbers are 27 and 16. 


When we have two unknown quantities, we must always 
have two equations. It will be seen that if we did not know 
the difference of the two numbers above, but only that x + y 
= 43, the numbers might be 40 and 8, or 23 and 20, or any 
others that amounted to 43. When we also know that x 
—y = 11, we can find the numbers to be 27 and 16, as 
above. 

The two equations must also be independent. Thus, if 
they were « + y = 43 and 2x + 2y = 86, we would not 
know what the numbers were. They might be any two 
numbers that amounted to 43; for twice that amount would 
be 86. The second equation would not be independent of 
the first; for it could be produced by multiplying both 
members of the first by 2. 

When we added the equations « + y = 48 ae x —y 
= 11, we produced the new equation 2x = 54, which con- 


88 ELEMENTARY ALGEBRA. 


tains but one unknown quantity, z. Thus we eliminated 
the other unknown quantity, y. By subtracting, we might 
have eliminated x. Thus. 








x+y = 43 

e—y=tili 

Subtracting equals from equals, 2y == 32, 
f= 16; 


The result is the same as before. The method of elimina- 
tion by addition or subtraction will.now be further explained. 


PropiEmM 2. Given the equations 1a ay 3 s 6 i 


to find the numbers which & and y stand for. 


Sotution. In this case, adding or subtracting does not 
eliminate either x or y, because the coefficients are different. 
Adding, we would have 8a + y = 44; subtracting, we 
would have —2% + 7y = 18. Neither x nor y would be 
eliminated. When this is the case, the equations must be 
prepared for elimination. Multiply both members of the 
first equation by 5, and both members of the second equa- 
tion by 3. We shall then have, 

Multiplying by 5, 15a + 20y = 155. 
Multiplying by 8, 1lba— 9y= 389. 

Now the coefficients of x are the same, and by subtract- 
ing, we may eliminate x. It is easily seen that we must 
subtract. By adding, we would get 30” + Illy = 194, so 
that there would be no elimination. Performing this oper- 
ation of subtraction, 


15a” + 20y = 155. 
15a— 9y= 39, 


Subtracting equals from equals, 29 y = 116. 
y= 4. 


ELEMENTARY ALGEBRA. 89 


Putting 4 times 4 instead of 4 in the first of the original 


equations, we have 32+ 16 = 31. 
Transposing, 32 = 31 — 16. 

5 ra a 

Eth e 


Hence, x = 5 and y = 4. 


In the preceding solution, we multiplied the members of 
the first equation by 5, which was the coefficient of # in the 
other equation. We multiplied the members of the second 
‘equation by 3, which was the coefficient of x in the other 
equation. We can always make the coefficients of a letter 
the same in this way. If the coefficients have a common 
factor, it is best not to take the trouble of multiplying by 
that, but by the other factors only. 


: 32 +2z= 16 
PrRoBLEM 3. Giyen ee egrets 30 f to find the 
values of x and z. 
SoLuTion. It is easily seen that if we multiply the mem- 
bers of the first equation by 3, the coefficients of x will be 
the same. 








Multiplying by 3, 9x + 6z = 48. 
Leaving the second equation un- 
changed, 9x2 — 8z = 20. 
Subtracting equals from equals, 142 == 28, 
Zu 2, 
Putting 2 times 2 instead of 2 z in the first equation, we 
have 382+4=—> 16. 
Transposing, 3x2 = 16 —4, 
50 == 12, 
cea 4, 


Hence x = 4 and z = 2. 
8 * 


90 ELEMENTARY ALGEBRA. 


PROBLEM 4, Given | Oar 4 ax o ; to find « and y. 
Sotution. Putting 4 times 4 instead of 4y in the first 
equation, we have 82+ 16 = 25. 
3x = 25 — 16. 
S@= YQ, 
c= 3, 
Hence x = 3 and y = 4. 
PRoBLEM 5. Given hes ely area ; to find x and y. 


Cine s ated 


EXPLANATION. Here, if we multiply both members of 
the second equation by 4, it becomes the same as the first. 
Hence by adding we would get 82 + 8y = 56, and by 
subtracting we get 0 = 0. There is no elimination. This is 
because the equations are not independent of each other. It 
is just the same as if we had but the one equation x + y = 
7; for then we could multiply both members of it by 4 and 
get the other. Hence any numbers which amount to 7 
would do for the values of x and y. The problem is inde- 
terminate. 


EXAMPLES. 


Bt to find x and y. 
Ans, 2 = 639 =a 

i} to find x and y. 
Ans. x = 8; y = 4. 


a to find # and y. 


Ans. & = Ot eae 
ie to find x and y. 


+4 
bo 
<&S 

I 


Ans, = 33 =o 


ELEMENTARY ALGEBRA. 91 


le ff to find x and z. 

ADgi dT ea ei an 
Mota g { to find a and 2. 

Ans. «© = 13; 2 = 4, 


y—2x2=1 
Ye = 5 to find « and y. 
Anse Go == ly yi=s5. 
. 2y—xz#=7 
8. Given ey, ; to find x and y. 
Ans. £ == 3; y= 8: 
17y —15x = 49 
9. Given Hoe 37 | | to find 2 and y. 
Anat se — Ls yaa 
10. ee fee, & 7 tt find x and y. 
Ang 2 = 2-47 = — EE 


Nore. The members of the first equation can be divided by 11 
and the equation reduced before eliminating. 


17y + 5x2 = 101 


11. Given | Goh maae 


e to find x and y. 


Ats. 0 == 1h). ia 


ig en BY 


. Given } 16% 4 26y — 162 


to find z and y. 
Ans. © = 2; y= 5. 


3} 
13. Given | a = Bea \to find y and z. 
ANS -Y == 10 (2 =a 
(4. Given 17 Fiat ie he iy to find @ and y, 
ANG. Disa) ant 
i5. Given cite 23 tare 3 | to find x and y. 


ANS, es 2 epee 


92 . 


16. 


Ve 


18. 


19. 


20. 


en 


22. 


23. 


24. 


25. 


26. 


27. 


ELEMENTARY ALGEBRA. 


1342+ 2y= 180 
doe ED 192 ¢ # find x and y. 
Ans. « = 12; y =z 


Given Agee ane ar 335 f 1 find x and y. 
Ans. © = 2; y= 9. 
| to find x and y. 


Ans. x =1; y = 60, 


89 
291 


ae 
bo 
© 
8 

ll il 


aa 
+ 
Ans, «= 1934) eee 
Given ay nf iy mii 36 ¢ find x and y. 
Ans. © = "15 Yee 
a 


ai 
a 
Ans. © == 2s yee 
ri 
a 


Given | ,* "Y= fo { to find « and y. 
Ans. w= 7; y=. 
Given 37 3, = at to find @ and y. 
Ans. 2 = 5; y = 2. 
Given O71 gy = — 19 f tofind « and y. 
Ans. © = 38; y= 58. 
Given } 457 t 99% Bi eerie find x and y. 
Ans. «x = 10; y= 9. 
Given ce a 5 to find x and y. 


Ans. 2 = 174 


28. 


29. 


30. 


dl. 


32, 


33. 


34, 


35. 


36. 


37, 


38. 


39. 


ELEMENTARY ALGEBRA. 93 


. 1091 « — 576y = — 969 

Given J _ 10912 + 596y = 889 to find x and y. 
Ans. «= —38; y = —4, 

° G—y = § 

Given 12 9 = 6 | to find a and y, 


Ans. m= 4; y= —4, 
; 2 3y = 29 
Given J rt 4y = 93} to find «and y, 
Ans. = 4; y= 7. 
Given oF TY = 28) to find andy 
Ans. c= 9; y= 9, 
. 2” —32=—15 
Given | Re 57 to find x and z. 
Ans. «= 3;2= 7. 


Given Dnt 5s = 31 to find @ and z 
Ans. x = 2; z= 5, 
Given } eels “40 ft find x and w. 
Ans. © = 20; w= 1. 
Given lates Pe to find x and w. 
Ans. & = 11; w= 22. 
Given fut y= to find w and y. 
Ans. w= 2; y = —2, 
33 to find p and q. 
Ans. p= 4; q =4, 
ee 
Ans. § = —4; t= 4, 
Given ee hee hs to find s and ¢. 
Ans. s = 100; ¢=1. 


94 ELEMENTARY ALGEBRA, 


SECTION XXIV. 


Questions producing Equations with Two 
Unknown Quantities. 


ProspLEM 1. One man buys 5 collars and 7 handker- 
chiefs for $3.80; another buys at the same rate 12 collars 
and 2 handkerchiefs for $3.20. How much are the collars 


and handkerchiefs apiece ? 


Soutution. Let x represent the price of a collar, 
And y the price of a handkerchief. 
Then we have, by the statement of the question, 
5a + Ty = 380. 
122+ 2y = 320. 
By solving these equations, it will be found that the col- 
lars were 20 cents each, and the handkerchiefs 40. 


PROBLEM 2. A person has ten-cent notes and twenty- 
five-cent notes, 100 in all, whose value is $13. How many 
has he of each? 


SotutTion. Let x = the number of 25-cent notes he has, 
And y = the number of 10-cent notes he has. 
Then, by the sas of the question, 
ee: 
25 2 v 10y = 13800. 
By solving these, it will be found that he had 20 twenty- 
five-cent notes and 80 ten-cent notes. 


EXAMPLES. 


1. A gentleman bought an encyclopedia in 24 volumes, 
and a history of England in 6 volumes, for $132. The en- 
cyclopedia cost $3 more per volume than the history. What 
was the cost of each per volume? Ans. The encyclopedia 
cost $5 per volume, and the history $2 per volume. 


ELEMENTARY ALGEBRA, 95 


2. A merchant has some wine worth $5 a gallon, and 
some worth $10 a gallon. He wishes to make a mixture of 
20 gallons worth $7 a gallon. How much of each must he 
take? Ans. 12 gallons at $5, and 8 gallons at $10. 

Notr. The whole 20 gallons will be worth $140. 


3. How many gallons of wine at $1.25 and at $1.00 must 
be mixed to fill a cask of 63 gallons at $1.10? Ans. 251 
gallons at $1.25, and 374 gallons at $1.00. 

4, Find two numbers whose sum is 27, and their difference 
3. Ans. 15 and 12. 

5. A hatter sells to one person 7 hats and 11 caps for $39; 
and to another, at the same prices, 5 hats and 1 cap for $21. 
What was the price of each? Ans. $4 fora hat, and $1 for 
a cap. 

6. At what price are horses and oxen sold, when 4 horses 
and 5 oxen cost $300; and 9 horses and 10 oxen cost $650? 
Ans. $50 for a horse, and $20 for an ox. 

7. Eight Spanish doubloons and ten pistoles are worth 
$163.12. Two doubloons and five pistoles are worth $50.49. 
What is the value of each? Ans. A doubloon is worth 
$15.535, and a pistole, $3.884. 

8. Find two numbers such that their sum is 34, and five 
times the greater minus three times the less is 58. Ans. 20 
and 14, 

9. Find two numbers such that double the greater minus 
the less is 22, and double the less minus the greater is 1. 
Ans. 15 and 8. 

10. Divide 38 into two such parts that one may exceed 
the other by 6. Ans. 16 and 22. 

11. A and B together own property amounting to 
$120,000, and seven times*A’s property is equal to five 
times B’s. How much has each? Ans. A has $50,000, 
and B has $70,000. 


96 ELEMENTARY ALGEBRA. 


SECTION XXV. 
Factoring. Fundamental Theorems. 


Theorem I, The square of the sum of two quantities is 
equal to the square of the first, plus twice the product of the 
first and second, plus the square of the second. 


a+ 06 

Proor. Let a and 6b standforany a + 6 

two quantities. Then, multiplying a a 4 The 
+ b by itself, we have what the theorem ab 4. # 


states. ————___—____—_— 
v+ 2ab+ B 


Theorem II. The square of the difference of two quantities 
as equal to the square of the first, minus twice the product of 
the first and second, plus the square of the second. 


a— 6b 

Proor. Let a and 6 stand for any a — 6b 

two quantities. Then, multiplying a (2 op 
— 6 by itself, we have what the theorem __. ae 
states. _t 
a? —2ab + 0 


Theorem III. Zhe sum of two quantities multiplied by 
their difference is equal to the difference of their squares. 


a+ 56 

Proor. Let a and 6 stand for any a — 6b 

two quantities. Then, multiplying a at sabe 
+ b by a —), we have what the theo- _2 oh aoe 
rem states. 2 A 


Theorem IV. The difference of the same powers of two 
quantities is always evenly divisible by the difference of the 
quantities. 


ELEMENTARY ALGEBRA. 97 


Thus, we find by performing the operation of division, 
(a? —b*) ~ (a—b) = a+ Bb. 
(a? — 6°) + (a —b) = a’ + ad + OB. 
(at — bt) + (a —b) = a® + ad + ab? + OD. 
(a —b’) + (a —b) = a + a + ad’ + ad? + OF. 
(a —b°) + (a—b) = @ + ad + ab’ + a’? + alt 


Any other expressions might be used instead of a and 6. 
Thus, we may find in the same way, 
(mn? — c?) + (mn —c) = mn + ¢. 
(m'n3 — c®) + (mn —e) = m’n’? + mne + ©. 
(mint — ct) > (mn —c) = min? + m’n’e + mne’ + &. 
&e., &e. 
Also, ( 9 —4) + (8 —2) = 34 2. 
(27 —8) + (8 —2) =9+644, 
&e., &e. 


The complete proof of the theorem is too difficult for insertion 
here. As far as the student needs to use it, he can prove it by per- 
forming the operation. 


EXAMPLES. 


Write down, where you can, the values of the following 
expressions, without ‘actually performing the work. State 
what theorem you use in obtaining your result. 

Thus: What is (x + y) (« —y) equal to? Ans. To 2 
—y’, by Theorem III. The sum of two quantities multiplied 
by their difference is equal to the difference of their squares. 


Note. Let the student tell why parentheses are needed wherever 
they are used in this section. 


1. (e + dy’. Ans. 2 + 2ed +. a’ 

2. (e —d)’. Ans. ¢ —2ed + d’. 

3. (a + 1)”. Ans. @ + 2a4+ 1. 

4. (ec — x)’. Ans. ce? —2cx + x’. 

5. (1 +. y)*. Ans. 1+ 2y+ y’. 
9 G 


ELEMENTARY ALGEBRA. 


6. (ab + c)?. Ans. ab? + 2abe + @’. 

7. (8@¢@ + a)’. Ans. 25 at + 10 a*x + a. 
8. (8¢+ 2y) Ans. 9 + 12 cy + 4y’. 

9. (ed + y’)’. Ans. ced? + 2cdy’? + y'. 
10. (ex — 3 y’)’, Ans. a? —6 ery? + 97. 
11. (Sax + 37°)’. Ans. 25 atx’? + 30 a@zry> + 97. 
12. (6cedy + 208’). Ans. 36 e'd*y’ + 24 ab’cedy + 4 a’b*. 
13. (2 + 3) Ans. 4+ 12+ 9 = 25. 
14. (6 —5)’. Ans. 36 —60 + 25 = 1. 
15. (6¢ — 0)’. Ans. 36c? —0 + 0 = 362. 
16. (3 ab + ed)’. Ans. 9 ab? + 6abed + ed’. 
17. (4can + 3m’ar’y’)’. 


Ans. 16 c?n?x? + 24 em’nx*y? + 9 mitxty*. 


18. (ab — ed)’. Ans. ath? —2 abe'd + cd’. 
19. (52+ — 6 2°), Ans. 25 2§ — 602" + 362% 
20. (ce? + 2ax). Ans. ct + 4ac’a + 47x”, 
21. (a + ©) (a—*2). Ans. a? — x’, 
22. (ec —y) (c+ y). Ans. ? —y’. 
23. (2a + 36) (2a —3D). Ans. 4a? —9 6. 
24. (5m.+ x?) (6m —2?). Ans. 25m? — x*. 
25. (8a + c) (8a—ce). Ans. 64a? —c’. 
26. (3 abet — 5 mn®) (8 abc! + 5 mn’). 


Ans. 9 a*‘b’c® — 25 m?n™. 


27. (Qaminx + y) (Ya'mnx —y). Ans. 81 atm'n’?x? — y’. 
28. (5 —3) (5 + 3). Ans, 25 —9 = 16. 
29. (6 — 2) (6 + 2). Ans. 36 —4 = 32. 
30. (9¢ + 1) (Qe —1). Ans. 81 c? —i. 
31. (8ab + 0) (8ab —0). Ans. 64 a’b? —0 = 64a7b”. 
32. (6y + 2am) (5y —2am). Ans. 25 y? — 4a?m’, 
33. (5 + 4) (5 —4). Ans. 25 —16 = 9. 
34. (6 + 4)”. Ans. 36 + 48 + 16 = 100. 
35. (6 — 4)’. Ans. 36 —48 + 16 = 4. 
36. (m® —n*®) + (m —n). 


Ans. m' + mn + min? + min? + mint + mn? + mn® 


+ n. 


ELEMENTARY: ALGEBRA. 99 


37. (c& — b*) + (ec — bd). 
Ans. & + eb + cb? + cb? + cbt + 0B. 
38. (7° —s*) + (r —s). Ans. rv? + rs + 8”. 
39. (mi —r") + (m —7r). 
Ans. m® + mr + mtr? + mr3 + mrt + me 4+ 
40. (a7b? — x’) + (ab — 2). Ans. ab + x. 
41. (ea? — y®) + (cx — y). 
Ans. cat + ea®y + ea*y? + cxy® + y'. 
42. (a*b* —c*d*) + (ab — cd). 
Ans. ab? + ab’ced + abe’d? + ed’. 


43. (a? —1) + (a—1). Ans. a + 1. 
44, (at —1) + (x —1). Ans. vo + 2? +2 +4 1. 
45. (1 — ec) + (1 — 0). An. 1l+e+e4+e. 
46. (25 —9) + (5 —8). Ans. 5+ 3= 8. 


47. (125 — 27) + (5 —83). 

Ans. 25+5x3+9 = 49. 
48. (a® — B®) + (a —D). 
Ans. a® + ab + a®d? + add? + att + a®d® + a’! 
+ ab’ + BD. 

49, (m® — x*) + (m — 2). 
Ans. m® + mtx + mia? + m2? + mat + 2. 

50. (g* — cht) + (g —ch). 
Ans. g& + gch + gch? + &h’. 


QUESTIONS ON THE FOREGOING SECTIONS. 


What is the difference between (a + xz) (a —zx) anda -+ za —z, 
oratzxXa—zx? 

Is a + xa monomial, a binomial, or a trinomial? Why? 

Is a binomial a polynomial? 

Write three monomials, each containing four letters and a number. 
Write three binomials, three trinomials, three polynomials of seven 
terms each. 

Name all the coefficients you have just written. 

What is the difference between addition in arithmetic and addition 
in algebra ? 

How is any term subtracted from another ? 


100 ELEMENTARY ALGEBRA. 


What is the rule for the signs + and — in multiplication and di- 
vision ? 

How do you multiply together powers of the same letter ? 

How do you divide one power by another power of the same letter? 


SECTION XXVI. 
Factoring. Prime and Monomial Factors. 


A measure or factor of a quantity is anything which di- 
vides it without a remainder. Thus, 5 isa measure or factor 
of 15; ais a measure or factor of a’ or of ax; ¢d’e is a fac- 
tor of cd’ex. 

A prime quantity is one which has no factor but itself and 
unity. Thus, a, 2,7, and a + & are prime; but az is not, 
as it has a and « for factors. . 

The prime numbers between 1 and 50 are 2, 3, 5, 7, 11, 
1O2 14,19, 20;-20; 01, Olo oleae ys 

The prime factors of a are a and a. 

The prime factors of 24a’c’d are 2, 2, 2, 3, a, a, ¢, ©, ©, 
and d,. 

The prime factors of 7‘ are y, y, y, and y. 


Polynomials, it must be remembered, are expressions 
composed of two or more terms, connected by + or —. 
They often have a monomial factor which may be separated 
from the rest. 

Thus, in the polynomial ax + a’, both terms may be di- 
vided by the monomial « Performing the division, we get 
a+ <2. Hencethe factors of aw + x? arexanda+z2x. In 
other words, av + 2? = 4 (a+ @). 

Again, in the polynomial 24 a’cx — 36 abe’x’ 4. 8 a’ex%, 
each term may be divided by 4, by a’, by ¢, and by «. That 
is, the monomial 4 a’cx is a factor of the polynomial. Per- 
forming the division, we get 6ct—Qac’x + 22°. Hence 


ELEMENTARY ALGEBRA, 101 


the polynomial is the same as 4 a’ex (6 ct —9ac’a 4+ 22”), 
The parenthesis is used to show that each term within it is 
multiplied by 4 a’ex. 

In the same way we may find that the factors of 35 a‘x 
—2ax+ x” are « and 38a*—2a+ 2. In other words, 
38a%—2JZaxr+2?=—-2 (8Bat—2a+ 2). 


EXAMPLES. 


1. What are the prime factors of a°? Ans. 4, @, a, a, a. 
2. What are the prime factors of a*a*y? 
Ans. 4, Q, A, £0, 2, Yy. 
3. What are the prime factors of 5 a*x*y? 
Ans, .5,&, 0; 0, X,Y. 
4. What are the prime factors of 24 a’z’? 
Anse 352522, a, a, @, @, G;.0;.0;,%,2. 
5. What are the prime factors of 18 m‘n? 
Ans. 3, 8,2,m,m,m, m,n. 
6. What are the prime factors of 21 p*q’x? 
Ans. 3,7, ),D; D5, J, 2: 
7. What are the prime factors of 135 abbx’y ? 
Ans. 5, 3,3, 3, a, a, a, a, a, 6, X, X, ¥. 
8. What are the prime factors of 235%’? Ans. 5, 47, k, k. 
9. What are the prime factors of 37 ? 
Ans. 37 is itself prime. 
10. What are the prime factors of n? 
Ans. n is itself prime. 


Separate each of the following polynomials into two fac- 
tors, one of which is the greatest monomial factor in the ex- 
pression. 


11. 2a+ 2-2. Ans. 2 (a + x). 
12. 13a—138. | Ans. 13 (a —6). 
13. 24a? + 24. Ans. 24 (a? + 1). 


14. 32 a® + 48 a®b — 24 ab’. 
Ans. 8a’ (4a? + 6 ab — 30’). 
15. 5a + 10y. Ans. 5 (a + 2y). 
Q* 


102 ELEMENTARY ALGEBRA. 


16. 14ac —7d. Ans. 7 (2 ac — d). 
17. ab — ax, Ans. a (6b —2). 
18. ax + Qay. Ans. a(x + 2y). 
19. 10 cy — 10 xz. Ans. 10” (y —2). 
20. 10 xy + 15 xz. Ans. 5x (2y + 32). 
21.) ab — 20 ar, Ans. 5a (b —5 2), 
22. ab + a. Ans. a (6 + 1). 
23. ab + 0. Ans. b (a + 1). 
24, 30 ab — 54 b. Ans. 6b (5a — 9). 
25. 25 a’y — 25 az. Ans. 25 @ (y —2). 
26. 3a + 3. Ans. 8 (a + 1). 
27. 840° — 34 a5. Ans. 34a° (a —1). 
98. 27 m? + 9 mn. Ans. 9m? (3 + n). 


29. 477 —16 7°s + 2475s? 4+ 12 7°s°, 
Ans. 47° (1 —48 + 67°s’ + 31°s*). 
30. 8aty + 8a’y — 18 ay’ + 33 aay. 
Ans. ay (8a + 3—18y + 38 2a’). 
31. 36 b'z — 21 ba — 9 bY 2% y’. 
Ans. 3 6° (12 b’2 —7 « —8 b’a*y’). 
82. 15 am?yz? — 12 b'm'n?y? — 9 a*b?min’. 
Ans. 8 m? (5 abyz’ — 4 b?mn y? — 3 a7b’m'n’). 
33. 216 a®btetdéa®y — 96 aed®x’. 
Ans. 24 a’ctd’x*® (9 abty — 4c). 
34. 56 ad?’mniatz + 224 bd'mxz. 
Ans. 56 d?maz (a’e?m®n'a? + 4 bd*2*). 
39. 216 m’n®p’q’s + 36 m?n'p’*qs’. 
Ans. 86 m’n®p’qs (6 mq’ + 8°). 
36. 185 a?y?z? — 87 a®bed?y’z. Ans. 87 a’y’z (5z —abed’). 
37. 25 ate?x*yt2® — 25 abe’aty'z5. Ans. 25 a®c’x*y*2’ (a — 2). 
38. 12 a®e’atyz® — 5 ale’atyz'. Ans. aeatyz (12 a — 9). 
09. 8 ab’e’+ 5 ac’d —5ad’e + 8 ae’*f. 
Ans. a (8b'e + 5d —5d’e + 8 ef). 
40. 2ay + 2y? + 2yz. Ans. 2y (a@ + y+ 2). 
Nore. The foregoing examples may be proved by reversing the 
operation, that is, by multiplying the factors together. 


ELEMENTARY ALGEBRA. 103 


SECTION XXVIII. 


Factoring. Applications of the Fundamental 
Theorems. 


PrRoBLeM. Factor r? + 27s + 8’. 


Soxtution. The first term is the square of 1, the last term 
is the square of s, and the middle term is plus twice the 
product of rand s. Hence, by Theorem L., it is the square 
ofr+s. Thus, 7? + 2rs+ 3? =(r+s)% 


PrRospLEM. Factor 4n? —12n + 9. 


Sotution. The first term is the square of 27, the last 
term is the square of 3, and the middle term is minus twice 
the product of 2m and 8. Hence, by Theorem IL., it is the 
square of 2n —3. Thus, 4n? —12n + 9 = (2n — 3). 


ProspuEM. Factor 25 x*/* —4n’. 


Soturion. The first term is the square of 5 x’y’, and the 
second term is minus the square of 2. Hence, by Theorem 
IIL., it is equal to the sum of 5 x’y? and 2 multiplied by 
their difference. Thus, 25 aty8 —4n? = (5 2’y® + 2m) 
(5 a?y® —2n). 


It is an excellent exercise to tell of a trinomial whether 
it is a square or not, by observing whether or not Theorem 
I. or Theorem II. applies to it. 

Thus, 25 a? + 40 az + 16 2’ is the square of 5a + 42, by 
Theorem I.; and 25 a* — 40 az + 16 2’ is the square of 5a 
—4z, by Theorem II.; but 25 a? — 20 az + 162’ is not a 
square at all, as the middle term is not twice the product of 
5a and 42. 


In the following examples, when the trinomial is not a 
square, tell why it is not. 


104 


ELEMENTARY ALGEBRA. 


EXAMPLES. 


Tell which of the following trinomials are squares, and 
factor those which are. 


92? + 24 az + 16 27, 
N25 we 16 x2 + 9 23, 
. 8627? + 12 427-4 2. 

. 062? + 12424 42% 


9y + 80yz 4+ 22”. 
v+b6ab+ 90%. 


. 25m? + 20mn + 4n2, 
. 81 p? — 86 pq + 4¢q’. 


64 ¢? — 12 cr + 9 x’. 


. 640 —48 cx + 9 2? 

164 .cC'—o2e0 4+ 425 

. 36 & — 54ed + 49d’. 

. 067? — 84 rs + 49 8”. 

. 9a7bt + 21 a?b’exr + 49 a’e*z’. 
. 9a7bt —42 c7b’er + 49 ae?x”. 


. 144 mn? — 72 m?nx® + 9 mtx. 


. 121 p’s* + 32 r*ps* — 25 7’. 

. 81 mn + 33 m’?n? + 25 min, 
. ary? + 48 2°y'z + 64 a? yt”, 
. 9 a®y® + 82 a y'z + 36 x*y'2”, 
. 9 aby? + 86 a y®z + 86 a’y'2”. 
. 9 ay + 36 a y’z + 16 x’y'2”. 
. 9 ar8/h + 24 ay®z + 16 x’y*2”. 
. 9 xy? + 18 xyz + 9 xy*2?. 

. 9 xy? — 12 aryfz + 4 ax yt2?, 

. 64073 — 48 aca + 9 x”. 

. 64 act — 48 ac’a — 9 2”. 

. 64 ee + 48 ac’x — 9 x. 

. 64 act — 48 wey’? + 9x. 

. 64 ac —48 aca + 9 x”. 

~ Imint + 12 mint + 4 mint. 


Ans. (3x + 42). 
Ans. Not a square. 
Ans. (6x + z)’* 
Ans. Not a square. 
Ans. Not a square. 
Ans. (a + 36)’. 
Ans. (5m + 2n)?. 
Ans. (9p —2q)’. 
Ans. Not a square. 
Ans. (8e —382)*. 
Ans. (8¢ —2«)’. 
Ans. Not a square. 
Ans. (6r — 7s)’, 
Ans. Not a square. 
Ans. (3 ab? —7 acx)’, 


Ans. (12 mn — 3 m’2°)?, 
Ans. Not a square. 
Ans. Not a square. 
Ans. (3 xtyt + 8 xyz)”, 
Ans. Not a square. 
Ans. (8aty* + 6 ay’z)’. 
Ans. Not a square. 
Ans. (3 xty* + 4 xyz). 
Ans. (3 aty* + 3 xy’z).. 
Ans. (3 xty* — 2 xy’z)’. 
Ans. Not a square. 
Ans. Not a square. 
Ans. Not a square. 
Ans. Not a square. 
Ans. (8 ac? —32x)?. 
Ans. (3 mn? +- 2 m?n?)*. 


32. 
30. 
34. 
39. 
36. 
ov. 
38. 
og. 
40. 


ELEMENTARY ALGEBRA. 


8a? + 16ab + 402 


mn? + 2mn + 1. 
mn? + 2mn + 4. 
w+ 2Qae+ x. 
a’ —2Qax + x”. 
4a’? + 4ab + 0% 
4a’? —4ab-+ b* 
927+ 24x + 16. 


64a? — 80a -+ 25. 


Factor the following: 
41. 
42. 
43. 
44, 
45. 
46. 
47. 
48. 
. 9at — 25 Dt. 

% a} — gh, 

: a? _ q?®, 

. wa? — §”, 
ciao. 20 0”, 
- 42° —162t . 
. 16 a*b’c*d? — 9. 
. 81 — 86. 

. 64— a’. 

. 2? — 49. 

. 9 —4. 

. at — cde’, 


Note. 


a —2Z’, 

A9 a? — 64 y’. 
4y’ —9 2", 

16 2’ — y%. 

25 x — 6", 

64 a& — 9 yx. 

9 ee ues 4 Ce ah 

25 a’® — 49 b8y¥, 


105 


Ans. Not a square. 
Ans. (mn + 1)?. 
Ans. Not a square. 
Ans. (a + 2), 
Ans. (a —x)’. 
Ans. (2a + 6)*. 
Ans. (2a —b)’. 
Ans. (3x + 4)’. 
Ans. (8 a—5)*. 


Ans. (a + x) (a — 2). 

Ans. (Tx + 8y) (7x —8y). 
Ans. (2y + 32z) (2y —32z). 

Ans. (42 + y’) (4z—y’). 

Ans. (5 2*® + 68) (528 — 6°), 

Ans. (8a + 3y’) (8a —3y’). 
Ans. (3 y® + 2 2q’) (3 y® —2 zq’). 
Ans. (5 a® + 7 b’y") (5 a —7 By’). 
Ans. (3a + 56?) (8 a? — 5D’). 
Ans. (a5b® + a®b*) (a®b> — a®d®). 
Ans. (a® + a’) (a® —a’). 

Ans. (a® + b*) (a® — B°). 

Ans. (5 a® + 56°) (5a® — 50°). 
Ans. (2x* + 427) (2a*—42’). 
Ans. (4abed + 3) (4 abed — 3). 
Ans. (9 + 6) (9 —6). 

Ans. (8 + a) (8 —a). 

Ans. (x + 7) (x1 —T). 

Ans. (3 + 2) (38 — 2). 

Ans. (a? + e’d'et) (a? — c'd'et). 


The foregoing examples may be proved by reversing the 


operation, that is, by multiplying the factors together. 





106 ELEMENTARY ALGEBRA. 


SECTION XXVIII. 


Factoring. (Continued.) 


WHEN it is desired to separate any algebraic expression 
into factors, it should be first divided by any monomial 
which will divide each of its terms; then by any other ex- 
pression which by the preceding four theorems, or by ob- 
servation, may be seen to be contained in it. 

Thus, taking the expression 28 bex’ — 84 bexy + 63 
y’be, it may easily be seen that every term is divisible by 7, 
every term contains b, and every term contains c. Hence 
we may divide by 7 bc, decomposing the expression into 
Tbe (4a? —12 ay + 9y’). The last of these factors is 
seen by Theorem II. to be the square of 2a —3y. Hence, 
finally, the original expression is equal to 7 be (22 —38 y) 
(2x2 —3y). 

Again, taking the expression, 17 a’ —17 a’y’, we may 
divide each term by 17 a’, decomposing the expression into 
17 x? (a® —y’). By Theorem IV., the last of these factors 
is divisible by « —y. Performing the division, we have, 
finally, 17 a? (wv — y) (at 4 ay + ay? + ay? + y'). 


EXAMPLES. 


Separate each of the following expressions into factors, 
one monomial, and the polynomial ones factored as far as 


possible. 
1. 9a —92’. Ans. 9 (a—x) (a+ &). 
2. 20 y? —45 2. Ans. 5(2y —32) (Qy+ 32). 
3. 1762 —11 y’. Ans. 11 (427 — y) (42? + y). 
4, 52 mt — 1300 nt Ans. 52 (m? —5 n’) (m? + 5n’). 
5. 822 —8 x’. Ans. 8x (2—2x) (2+ z). 
6. 19 a°b — 19 ab’. 


Ans. 19 ab (a? + 6°) (a + b) (a —B). 
7. 25 ab — 25 bt. Ans. 25 b (a —b) (a + ab + 0°). 


15. 


ELEMENTARY ALGEBRA. 107 


. sey —3 xy’. Ans. 3ay (x&—y) («© + y). 
. ar*® —2 ar’z + arz’. Ans. ar (r —2z) (7 —2). 
. dbx? + 10 Be’x + 5 bc’. 


Ans. 5 be’ (x + b) (a + b). 


Pigaea — 38 a'b’s* + 19 a7b.°. 
Ans. 19 a*b'x* (a —2x) (a — 2x). 
. dl aa —62a@bx + 31 ab ?x. 
Ans. 31 ax (a —b) (a —D). 
. das? + 10 atzt + 5 aPr', 
Ans. Sax (a + x) (a+ @). 
. d8ax’? —18 ax + 27 a. Ans. 3a (x —3) (a — 8). 
45 ay — 380 ay? + 5ay®. Ans. Say (8 —y) (8 — y). 
.387b4+ 18ye+ 9yxz Ans. 8y’? (b+ 6¢e4 822), 


. 8m’nx —18 mnx’ + 2mn'>y. 


Ans. 2mn (4mx —9 x? + nty). 


. 15 ay — 25 b’y + 1dc’y. 


Ans. 5y (80 —50B? + 3). 


. Bax® + 6an’y® + 42 ax’. 


Ans. 6 ax? (at + 4° + 7a). 


26 ax’ — 26 ay’. 


Ans. 26 a (« —y) (2° + ay + aty’? + a5y? + a’yt 
eye eh 


. 6a2t — 96. Ans. 6 (x? + 4) (a + 2) (x — 2). 
. 64 atd® — 25 a?b®. Ans. ab (8a —5b) (8a + 55). 
. 2477s —127°s*t —67rs. Ans. 6rs (4r —27’st —1). 


Soa —— Oy Ans. 5 (a? + y*) (fe + y) (« — y). 
. aent —atert. Ans. ate? (nr? + 1) (n + 1) (n —7). 
. ax? —a. Ans. a (« + 1) (« —1). 
at a2". Ans. a? (x + 1) (x —1). 
. a —a, Ans. a (a? + 1) (a + 1) (a—1). 
~la’y—22aey+ 11. Ans. 11 (zy —1) (zy —1). 


. 04am’? —- 17 atm + 51 a’mt. 


Ans. 17 a’m (2 atm — a? + 3m’). 


anche 


108 ELEMENTARY ALGEBRA. 


SECTION XXIX. 


Equations with Two Unknown Quantities. 
(Continued.) 


: 9x—2=10y41 
PROBLEM 1. Given [Fy 4 Say ode 


find the values of # and y. 


Sotution. In this case, transposition is necessary to 
bring the equations into the proper forms for eliminating — 
one of the unknown quantities. Transposing, we have 
9a — 10y 1 + 2. 

bye 3a—2xe4—y ob Ett 

| 92—10y= 3. } 
x+ 4y= 81. 

The easiest method of elimination here is to multiply the 
members of the second equation by 9, and subtract those of 
the first. Thus, 


Collecting, we have 


92+ 36y = 279. 


The first equation is, 9x—10y= 4. 
Subtracting, 46 y = 276. 
y= 6. 
Instead of 4y, put 24, in the second equation. Then we 
have, x+ 24 = 31. 
Transposing, x = 31 — 24. 


Hence x = 7 and y = 6. 


PROBLEM 2. Given 
[er uD + Dy oe 
22 +1 ey ee 
to find the values of x ae y. 


Sotution. In the first equation, performing the oper- 
ations of multiplication, we have 
xy + dy—Tx—385 = ay +y—I9Iaex—9+4+ 18. 


ELEMENTARY ALGEBRA. 109 


Here we cancel xy, by subtracting it from each member. 
Then the equations are 
nites 10 =y—9x—9+ a 
2¢+10=y-+4. 
Transposing, 
lama Sa 9x=— 9+ mt 
242 —y= 4—10. 
Collecting, and arranging, Sane ah ea as i 
Eliminating, and continuing the solution, it will be found 
tastes =), and y = 1(). 


: EXAMPLES. 
1. Given 


§ +182 —ay = (14—~2) (11+ y)+4+ 138 
12x+9y= 82+ 1l4y4+3 
to find the values of and y. Ans. x = 12; y= 9. 


2. Given ‘i 78 nw a aaah (y¥ +1) +5) to 


s 
find the values of and y. Ans. x= 4; y= 1. 

3. If $500 be taken from A’s property, and $1500 added 
to B’s, they will both have the same sum. If A’s property 
were five times as great as it is, and B’s six times as great 
as it is, they would together have $120,000. How much 
has each? Ans. A has $12,000, and B has $10,000. 


4, A number consisting of two digits is four times their 
sum. The number consisting of the same digits in an in- 
verse order is 18 more than the number itself. Required 
the first number. Ans. 24. 

Notre. Any number of two digits is 10 times the first, plus once 
the second. Thus 23 is 10 times 2, plus 3; inverting, it becomes 382, 
which is 10 times 3, plus 2. If z and y stand for the digits, 10% + y 
will be the number, and 10 y + z will be the number formed by in- 
verting the digits. 


5. If 21 years be added to the sum of A’s age and B’s, 


the result will be double A’s age. If 9 years be added to 
10 


110 ELEMENTARY ALGEBRA, 


their difference, the result will be double B’s age. Required 
their ages. Ans. A’s age is 36 years; B’s, 15 years. 

6. Find two numbers such that the greater minus 2 is 
double the less, and the less plus 22 is double the greater. 
Ans. 14 and 6. 

7. If A give B $100, B will have twice as muchas A. If 
B give A $200, A will have twice as much as B. How 
much has each? Ans. A has $400, and B has $500. 


8. If the greater of two numbers be multiplied by 5 and 
the less by 10, the sum of the products is 140. If the less 
be multiplied by 5 and the greater by 10, the sum of the 
products is 160. What are the numbers? Ans. 12 and 8. 


9. Ten years ago, A’s age was 10 times that of B. Six 
years hence it will be double that of B. Required the age 
of each. Ans. A’s age is 30 years; B’s, 12 years. 

10. Find a number consisting of two digits which is equal 
to 8 times the sum of the digits, the number consisting of 
the same digits in an inverse order being 18 more than their 
sum, Ans. 72. 


11. Find a number consisting of two digits which is 7 
times the sum of its digits, and is 27 greater than the num- 
ber consisting of the same digits in an inverse order. Ans, 
63. 


12. A grocer has two kinds of coffee. One pound of the 
first kind and one pound of the second are together worth 
48 cents. Four pounds of the first plus nine pounds of the 
second are worth $2.92. What is the price of each per 
pound? Ans. The first kind is 28 cents per pound, and the 
second 20 cents. 

13. A gentleman has two horses and two saddles. One 
saddle is worth $40; the other, $20. The best horse with 
the best saddle is worth $80 more than the worst horse with 
the worst saddle. The best horse with the worst saddle is 
worth $10 less than twice the value of the worst horse. 


ELEMENTARY ALGEBRA. 111 


What is the value of each horse? Ans. The best horse is 
worth $150; and the other, $90. 

14, A farmer, having 9 bushels of wheat at $2 a bushel, 
would mix with it rye at $1.40 a bushel and barley at $1 a 
bushel, so as to make a mixture of 50 bushels at $1.38 a 
bushel. How many bushels of rye and barley must he take? 
Ans. 25 bushels of rye, and 16 bushels of barley. 

15. Find a number consisting of two digits which is equal 
to 7 times the sum of its digits, and is equal to 9 more than 
12 times the difference of its digits, the unit digit being the 
less. Ans. 21. 

16. Find a number of two digits, the first being the 
greater, such that the number itself is 3 more than 7 times 
the sum of its digits, and 7 more than 16 times the difference 
of its digits. Ans. 52. 

17. Divide the number 149 into two such parts that the 
greater diminished by 5 may be double the less. Ans. 101 
and 48. 

18. A miller mixes wheat costing $1.25 a bushel with rye 
costing 75 cents a bushel. It amounts to 68 bushels, cost- 
ing in all $79. How many bushels of each kind does he 
take? Ans. 56 bushels of wheat, and 12 bushels of rye. 

19. Divide $3000 between A and B, so that A shall re- 
ceive a half-eagle as often as B does a dollar. Ans. A’s 
share is $2500; B’s, $500. 

20. Divide $5000 between A and B, so that 5 times A’s 
share shall be 4 times B’s. Ans. A’s share is $22222; B’s, 
$2777 4. 

21. A vintner wishes to fill a puncheon of 84 gallons with 
wine, so that it may have cost him $100.80 in all. He 
draws his wine from two casks which cost $1.00 and $1.42 
a gallon respectively. How much of each kind must he 
take? Ans. 44 gallons at $1, and 40 gallons at $1.42. 


22. I purchased 20 pounds of sugar and 25 pounds of 


112 ELEMENTARY ALGEBRA, 


coffee for $10; but the price of each having fallen 2 cents a 
pound, I purchased 29 pounds of sugar and 25 of coffee for 
$10.72. What was the first price of each? Ans. For sugar, 
20 cents per pound; for coffee, 24 cents. 


23. I bought 100 yards of linen and 150 yards of muslin 
for $82.50. I afterward bought 50 yards of linen and 200 
yards of muslin for $55, the linen costing 10 cents a yard 
more, and the muslin 5 cents a yard less, than at first. 
What were the first prices, and what the second? Ans. At 
first the linen was 60 cents, and the muslin 15 cents; after- 
ward the linen was 70 cents, and the muslin 10 cents. 


24. Find a number equal to 5 times the sum of its digits, 
such that if 9 be added to it, the digits will be inverted. 
Ans. 45. 

25. Find a number equal to 8 times the sum of its digits, 
such that if 45 be subtracted from it, the digits will be in- 
verted. Ans. 72. 


26. Divide the number 111 into two such parts that the 
difference between the greater and 120 shall be 9 more than 
50 minus the less. Ans. 86 and 25. 


27. A laborer engaged for 100 days, on condition that for 
every day he worked he was to have $1.25, and for every 
day he was idle he was to forfeit 50 cents for board. At 
the end of the time he was entitled to $55. How many days 
did he work, and how many was he idle? Ans. He worked 
60 days, and was idle 40 days. 


28. Divide 15 into two such parts that if the less be mul- 
tiplied by 5 and the greater by 7, the sum of the products 
shall be 93. Ans. 6 and 9. 


29. A gentleman owns two houses. If he should spend 
$900 in improving the better house, and $1000 in improving 
the other, the two would be equal in value. But if he should 
spend $2000 i in improving the better house, and $500 in im- 
proving the other, the first would be worth twice as much 


ELEMENTARY ALGEBRA. 113 


as the second. What is the value of each? Ans. One is 
worth $2000, and the other $1500. 


30. If a merchant mixes brandy and Port wine, putting 
in twice as much of the former as of the latter, the mixture is 
worth $22 a dozen. But if he puts in twice as much of the 
latter as of the former, the mixture is worth $24 a dozen. 
What is the price per dozen of the brandy and of the Port 
wine? Ans. The brandy, $20 per dozen; the wine, $26 per 
dozen. 

Nore. Three dozen of the first mixture are worth $66; of the 
second, $72. In other words, 2 dozen of the brandy plus one dozen 


of the wine are worth $66; and one dozen of the brandy plus two 
dozen of the wine are worth $72. 


SECTION XXX. 
The Least Common Multiple. 


A multiple of a quantity is that which contains it exactly. 
Thus 63 is a multiple of 7 and of 9. Also a is a multiple 
of a; xy is a multiple of x and of y; a? — 0? isa multiple of 
a —b and ofa + b. 

A multiple of two or more quantities is said to be a com- 
mon multiple of them. Thus az is a common multiple of a 
and 2; so are a’x, ax’, a’2*, a’x*, etc., common multiples of 
aand #; and az is the least of these. 

The least common multiple of several quantities is found 
in precisely the same way in Algebra as in Arithmetic: one 
or two examples will serve to show the process sufficiently. 


ProsieM 1. Find the least common multiple of 9 a’z, 
4 ax*, 12 ax’, and 6 a?z*, 
10* H 


114 ELEMENTARY ALGEBRA. 

















OPERATION. 
3 ) 9 ax 4 ax? 12 a®a? 6 a723 
2 ) 3a°x 4 ax? 4a°x” 2 a? 
2 ) Bae 2 ax? 2 ax? ax 
a ) I ee ax? ‘ih: a*a* 
a ) Sax oc or ax 
7 ) aa x a x? x 
x ) 3 ie ax Me 
3 1 a L 


Continually divide the quantities by any prime factor of 
two or more of them, setting down the quotients and un- 
divided numbers in a line below. When no two of them 
have a common factor, multiply together the divisors and 
final quotients. The product is the least common multiple. 

The product of 3, 2, 2, a, a, x, x, 3, a, and & is 36 a‘z*. 
Hence the least common multiple of 9 a’#, 4a2?, 12 a*a’, 
and 6 a’z? is 36 a’x’*. 


ProstEM 2. Find the least common multiple of 
(a+ x)’, a —2Qax + x’, a® —a’ax, and a? + a’x. 
OPERATION. 
a) (ata)? w@—2ar4+2? &b—aae ata 
a)(a+a) w@—2ar+2* a'—ar a+ae 
a+ax)(a+x)' a —2axr+a* a—s@ a+ a 
a— x ) atx @—2ar+2”? a—z 1 
a+ 2x a—x 1 1 
The product of a, a,a + %,@—%,a + #, and a —Z is 
a’ —2 atx? + a’a*. Hence the least common multiple of 
(a+), a —2 ax + 2’, a —a’a, and a+ a’e is a’ — 
2 ata? + axt, 














ELEMENTARY ALGEBRA. 115 


EXAMPLES. 


. Find the least common multiple of 9 2’, 63 a’x, 21 a'x%, 


14 ax’, 18 az, and 6 a’. Ans. 126 a’z*. 


. Find the least common multiple of aand b. Ans. ab. 
. Find the least common multiple of ad, be, cd, ac, ad, 


and bd. Ans. abed. 


. Find the least common multiple of a’, 6’c, ed, a’e, a’d, 


b’d, ab’, bc”, ed’, ac?, and ad’. Ans. a’b’c'd’. 


. Find the least common multiple of 3 and xy. 


Ans. 3 xy. 


. Find the least common multiple of a and a + 8. 


Ans. aw + ab. 


. Find the least common multiple of a’z, ax’, a? + a, and 


a+ x’. Ans. aba? + atat + abx® + a5, 


. Find the least common multiple of az + 2’, a + 2, @& 


— 2x’, and ax — x’, Ans. va —x', 


. Find the least common multiple of « — 3, 7 ay, « —4, 


and 7 y. Ans. T x®'y —49 a’y + 84 xy. 


. Find the least common multiple of 2? + y’, « —y, x 


+ y, and 2? —y’. Ans. ‘x — yf. 


. Find the least common multiple of r? — s’, (r —s)’, r 


—s,andr-+ s. Ans. 7? —7’s — rs’ + 8°. 


. Find the least common multiple of (a — x)*, a? —2 az 


+27,¢0—2’7,a+ 27, and a—z2. 
Ans. at —2a®x + 2 az’ — x, 


. Find the eet common mu iple of a? + Zax + 2’, 


(a+tax,anda+a. Ans. a + 8a’x + 8a2? + 23, 


. Find the least common multiple of 3 (a —b)’,4a —4, 


and 6 a —6b. Ans. 12a? — 24 ab + 12 6’. 


. Find the least common multiple of 24 (a —b), 12 ab’, © 


and 8ab. - Ans. 24 a’b? —24 a7b*. 


. Find the least common multiple of 9 a? —9 b?,4 (a+ b), 


AG + b)’,and4a—4ob. 
Ans, 360° + 36 a’b — 36 ab’ — 36 B*, 


116 ELEMENTARY ALGEBRA, 


17. Find the least common multiple of 15axz — 5 az’, 
(3 — ax)’, 15a —5aa, and 15 « — 52’. 

Ans. 45ax — 380 a2? + 5 az. 

18. Find the least common multiple of a? — x’, 5a? + 10 az 

+ 52’, v« —a’x’, and 5a’x', 

Ans. 5a’x*? + 5atxt — 5 a®x’® — 5 a? x*, 

19. Find the least common multiple of ab’ + abe, ab? — abc, 

and ab® — abe’. Ans. ab’ — abe’. 

20. Find the least common multiple of x —3, «7 —62 

+ 9,3b4—9b6,and3b. Ans. 3ba? —18 bx + 275. 


SECTION XXXI, 
Reduction of Fractions to their Lowest Terms. 


THE rules for operating on fractions should be proved in 
Arithmetic before the study of Algebra is commenced. 
Hence we shall generally only give examples of their appli- 
cation. | 
215 st wx 
265 stwaty? 

SoLutTion. W2 find by observation that 5, s‘, w®, and x 
will measure both numerator and denominator. Dividing 


them both by 5 s‘w*®x, we have pesca RIN which is the fraction 
53 s°a8y? 


PROBLEM 1. Reduce to its lowest terms. 


in its lowest terms, as nothing else will measure both terms. 


5 abc’ — 15 a7c® +- 20 a®ax* 
10 a’2° + 30 ax? — 20 ata? 


ProspueM 2. Reduce to its 





lowest terms. 


SoLution. We find by observation that 5, a’, and 2+ will 


ELEMENTARY ALGEBRA, 117 


measure both numerator and denominator. Dividing them 
gan0 6 $ 
both by 5a’x*, we have id ee Sa raledlelha Be nothing 
22° + 6ax — 4 a’x* 
else will measure both terms, this result is the fraction in its 
lowest terms. 


a + aa 


Peoriem 5. Reduce ———"~ 
aé + atx? + a’xt 


to its lowest 
terms. 


SotuTion. We find by observation that a? will meas- 
ure both terms. Dividing them both by a’, we have 
w+ 2 
at + aa? + xt 
asa—2,anda-+ x. It will be found that a —~2 will not 
measure either term, and a + a will measure the numerator 
only, giving as a quotient a?—ax-+ 2. Try that as a 
divisor of the denominator. It will be found to measure it. 
Then a? —ax + 2’ is a divisor of both terms. Dividing 
them both by it, we have __“ + __ which is the ori- 

at ax t+ x 


ginal fraction in its lowest terms. 


For further divisors, try such quantities 


Norg. If divisors do not readily occur to the pupil, let him try 
to factor both numerator and denominator. Then common factors 








may show themselves. Thus, ay mri when factored, becomes 
ax? — ay 
24/2 aig 
zy (4 —¥")> The common factor may be cancelled, giving a 
a (x? — y?) a 


EXAMPLES. 


Reduce the following fractions to their lowest terms. 


5 aX yz A 3 5a 
Chiara sige 

25 ata®y?2 Say 
eS atay P = mee 

12 min'p qe Ans 3 n'p’g 


28 m'naxy® ay 


118 


ELEMENTARY 


16 a’bictd*e® 
24 aSbctd’e? 
Dery z 
49a oytz 
106 a'mn?x* 





24 ma‘cx’yd 
18 p'a*baye 
121 abcde 

22 fghij 
35 x°y2°ep* 
45 may var yZp 











27 c’yzma 
36 2 yze'm 





Tac? + 21 atet — 5 abe® 





ae + dare? +12 act 
121 ad’ex + 55 ate? — 33 a? C 
11 ae 
34 Me — 12 ieee 














121 aed — 5atex? 
3aee + Qare’x —5atexr 








14axy —1l4axy + axy 


14axry —l4azxry + any 














x —y 
v—2xey+ y’ 
A oa ree 
a? —y? 
18,00 
3 + y® 
xy? —y! 


19. 


ALGEBRA. 


7 2 ax? 
" 3 minty 
4a’dma 
Ans. 308 pas 
11 abede 


Qfohi 
Ans. 7 ep 2 
9 mx 


Ans. 3 


4 a 
7 + 21 ave? — 5 ac 
ao + Dae 
ll@« + 5dac—3 
: =o 
Te — 6a? 
' 136+ 140% 
121 ed —5 ax? 


a 
e 


Ans. 





Ans 





Ans 





Ans 
Ans. 


1 
Ans. =. 
ns. i 
a: 
z—yY 
a+ wy ty 
xe+ey 
Ans. x — aly + + xy’ ayy 
— ea ty 


2 
Ans. ¥.. 


x 


Ans. 


Ans 


20. 


21. 


22. 


23. 


24. 


25. 


26. 


27. 


28. 


29. 


30. 


dl. 


32. 


33. 


ELEMENTARY ALGEBRA. 119 
































_. Jj Ans a baat 
a —xy + zy’? —y Se 
10a + 200% + 100° Ne 
15a* + 30 a5d + 15a°d® eo 
a’ — x" Ans Li 
at§ — a oe + 2x 
3 
are i yee las ga aut Hic 
xv —y a fe ARO Ree eat Sg i 
Seti iia Ana 
olkeoers iy a — aly + xy? —y 
Dp eg. Ans eee 
foo Bs 
rs —rs* r-+s 
oo 2 Sc EES Ans. cont eS 
rs — 3r°3? + 37r’s® — rst fas r —2rs + 3? 
xy — xy? Aneto 
—$———$ ———_—___.___.., ns. ° 
ay + 2aty* + xy? wate 
an” 7h ary Ane 1 
ax? — ary™ ie 
ee Ang een 
as—y’ aty 
$. peat al ;- Ans. The fraction 7s in its lowest 
a—2ax+2 
terms 
12 m‘n* —12 min? + 24 mn’? yer 2mn 
“18 mn? —18 mn? + 36 mn- bk 
ax + ay Anan 
Zax + ay 2x + y 
at y a Ans. - tips ae ASR ONS 
ab — x® | SY a ea 


120 ELEMENTARY ALGEBRA. 


SECTION XXXII. 


Reduction of Whole and Mixed Quantities to 
Fractions. 


PRoBLEM 1. Reduce ac + bx to the form of a fraction. 

Soutution. This can be done by merely placing 1 under 

ac + bx 
iia 


it as a denominator; for ac + bx is the same as 





ProsLEM 2. Reduce ¢ to a fraction having 6 as a de- 
nominator. 


SotuTion. A unit = - Then ¢ units are equal to ¢ 
b 
times ;-, or 3 Hence c = ae 


bi Sib b 


PRoBLEM 3. Reduce x + y to a fraction with the de 
nominator « — y. 





Sotution. A unit = ——# . Then x + y units = 








“—y _ ay 
eae. (a + y). Hence & + y = - EF, 


We see by the above reasoning that any quantity may be 
reduced to a fractional form by multiplying it by the pro- 
posed denominator, and placing the result over that de- 
nominator. 


PROBLEM 4. Reduce 54 to the form of a fraction. 


Sotution. Multiply 5 by 7, producing 35. Hence the 
5 = 3. Add 4 to %, and we have 39. Hence 54 = 32, 


PROBLEM 5. Reduce a + ~ to the form of a fraction. 


ELEMENTARY ALGEBRA. 121 


Sotution. Multiply a by n, producing an. Hence a = 


an ca, an an + ca 
=, Add = to Fi, and we have ——— 
n } 


PROBLEM 6. Reduce x —y — to the form of a 
fraction. s 


.Soutution. Multiply « —y by x + y, producing 2’ 


2 


—y’. Hence x —y= -—-—. As the fraction is nega- 





2 
tive, subtract ae from ¥. The operation of sub- 
dr Y y 


traction is this: “t—y 





sake}, ioe 2 
a5 os which is the same as —~ ee. : 


the minus sign before the whole fraction means fae each 
term of the numerator is to be subtracted. 


The result is 





, as 


— by 


Prope 7. Reduce — sgh) aes + y to the form of a 





fraction. 


Sotution. Multiply — ax + y by a — 6b, producing 
—ax + ay + bx — by. 
— ax + ay + bx — by 
aa a—b ; 
As the fraction is positive, add ant to this. Add the 


ba —2 
numerators. The result i is — oo ai ca ete Dey 





Hence — x + y = 


EXAMPLES. 
1. Reduce ac to a fraction whose denominator shall be cz”, 


3 2 
ackx 

Ans. “ 
Cc ex? 





11 


122 


10. 


dt, 


12. 


ELEMENTARY ALGEBRA. 


Reduce » + a to a fraction having the denominator 


n+ a. Ans 
n+a 
Reduce a? — 0’ to a fraction having the denominator 


a 0, Ans. 











Reduce x? + y’ to a fraction having the denominator 
: x 

ees Ans, C1 See 

Reduce n*x?y* to a fraction having the denominator 


an®a’y*t — en3x? 
a—c. Ans. UE panic ih! y* 
a—e. 





Reduce n’y’ —a’b? to a fraction having the denomina- 


2474, ,2 4},2,,2 

a’nty’ — atb’n 

tor n’a’. Ans. aga ; 
a 








Reduce n’y —a to a fraction having the denominator 





ney? pala 

8 eS 
n'y + a. Ans. ea 

Reduce x? + 1 to a fraction having the denominator 
at —] 

vee AL Ans. ee |: 


Reduce n? —a to a fraction having the denominator 
a’n? —a — n? + an 
a —n 








a? — n. Ans. 


Reduce 2° —y + ¢ toa fraction having the denomina- 


ax® —ax*y + ame 
tor az’. Ans. 
az* 











Reduce a@ + a to the form of a fraction. 





ab + ax 
Ans. ae. 


Reduce a + : to the form of a fraction. 


Ans 90a 
' c 


13. 


14. 


15. 


- 16. 


bi: 


18. 


19. 


20. 


21. 


ELEMENTARY ALGEBRA. 123 


Reduce ax + < to the form of a fraction. 





2,3 
Reduce ea? — — to the form of a fraction. 




















$8. as 
Ans. as sabes 
x 
Reduce m?n? — eae Y to the form of a fraction. 
Ans. — Be 
x 
Reduce ab — aes — to the form of a fraction. 
23.2 
Ans. Cs 
ed 
2 2 
Reduce a — x — in sol to the form of a fraction. 
a+ or 
22 
Ans. auita. 
Reduce a + «& + aed to the form of a fraction. 
204+ 2ar+ 2x 
Ans. ___ 
a+2zx 
2 2 
Reduce a + Rauk ree to the form of a fraction. 
a+b 
2 ab 
Ans. —— mau 
3 2 
Reduce ec + d — - ale : to the form of a fraction. 
vA he 
Ans. pres Eee 
Reduce 32? —2a + eet to the form of a fraction. 
2 
Ans. (hie, 


2 


124 ELEMENTARY ALGEBRA. 





























22. Reduce 3a — os to the form of a fraction. 
Ane ee 
a 
23. Reduce 5ab — i oe g to the form of a fraction. 
15 ab? —Ta —2 
Ans, — 3h —. 
- —8y7’ 
24. Reduce 82 —2y — eae Sy to the form of a frac- 
: 62°+ 47’ 
tion. Ans Sane j : 
25. Reduce 1 te i oa oe to the form of a fraction. 
28 
Ans. ao+e 
26. Reduce 1 — Z ee ze ile the form of a fraction. 
; 2 
Ans. a + BF 
27. Reduce ; + 1 to the form of a fraction. 
Ans aie Y, 
y 
28. Reduce * — 1 to the form of a fraction. 
Ans. See 
29. Reduce 1 — 2 to the form of a fraction. 
Ans. yee 
2 » 2 y 
30. Reduce — ee oe aa ed + 1 to the form of a fraction. 


a + a’ 
2a + Qax + 2at 


Ans. i. fae 





ELEMENTARY ALGEBRA. 125 


SECTION XXXIII. 


Reduction of Fractions to Whole and Mixed 
Quantities. 


2 2 
PrRoBLEM. Reduce a Boe Gg to a whole or mixed 








quantity. 
SotuTion. Divide the numerator by the denominator as 
far as possible. 
e—d)?’+ed+d@(e+2d 








ce —cd- 
Qed -+ d? 
2cd — 2d? 








3a 
Set down the quotient as a whole quantity, and if there 
is a remainder, set it down over the denominator. In this 
i he 
e—d 





problem the answer is ¢e + 2d + 


EXAMPLES. 


Reduce the following fractions to whole or mixed quan- 
tities : 





(Soon iaege Ans. weep ee 
a a 
24 

2. a a ; - Ans. n—m + f. 
2 

3. aad Ans. 6? + - 
2 ee 

eee rt i ee 
2 ltd a) 

5. a. Ans. 1. 


11* 


20. 


21. 





























ELEMENTARY ALGEBRA, 
SEG mo Ans. on ee 
3m 3 
25 mp — 9 mp R Ane Ae 
mp 
y Aare Fire 
18 aba —12 ax -5a Ane 8 bo ee 
6 ax 6x 
20 abe — 3 ae — 2e 38ac + 2e 
Ser a oe ee a} Ans. 4e¢ ——-: eos ih ° 
5 5 
7 +. Ans. x* —a*y + a’y’? —ay® + y'. 
3 3 
i Ans. x? —xy + y’. 
Bae. 8 2 2 5 ae 
Ty Ans. a® —a’b + ab? — B® + pe 
a’ + B i 2 
Aereny Ans. a Ot ae 
5 5 
aa Ans. xt + xy + xy? + xy’ + viet a 
dd — ed? Ans. ed — ed?. 
e+d 
wan Ange + 1. 
y y 
Somkd Ans. 1 — 4 
c c 
ao eRe Ans, 46120 + 2—© 
2ab + 26" Ans, 2b. 
a+ 6b 
a” + ax + a a 
eee Ans. a + ae 
ax + by + 2bx by + bx 
bisists ace Dia Ans Ce 








22. 


23. 


ELEMENTARY ALGEBRA, 137 




















2 2 ; 

oft 2ed + dl ees ht Some 

e—d Praha 

2° +ed+a@ -p 

ae Seige appa Ans. ¢ + aa Te 
a + 30% + 3800°+ Fh +e+4+d 
| Sg aera oe 

2 : e+d 

Ans. a? + 2ab + 6 ea 


m:* aT 2 mn +n? —m—n 




















i ee ee Ans. m+n —1. 
ary + axy’ + axy* | 
axy . Ans. 1 a y + ee 
a+t+ab+e : 
merED br Ans. a + wea 
a —ab —e ; 
Se Ry Ans. a — ee 
ety te + dry ty’ 

Bei. ane ie Anz. 1+u+y. 
ee 2 : 
: Es Ane oe ee 

x—y Ria 
br oe 5 aires 
a Ans. «* + ax + —[——<. 
Wie oy a is 
+ 
ve Ans. ax. 
an 
4 ctx —5 pele 
2 cx —2 Ans. 2¢? + ea: 
3 nht 
= es we Anahi ee 
ab a 
2 —— 
. ae Ans. axy + ay. 





128 ELEMENTARY ALGEBRA. 


SECTION XXXIV. 


Reduction of Fractions to others having the 
Least Common Denominator. 


a ac c 
Se2 ph ne ee ee and ite i ees to 
a*—c ac+e a” —ac 
equivalent fractions having the least common denominator. 


PROBLEM. Reduce 


Soxtutron. The fractions must first be in their lowest 
terms. The second is not so, both terms being divisible by 
c. Performing this, division, the fractions become 


a a q 
Seeley) ee . 
av—ce ate a” —ac 





We find the least common multiple of the denominators. 





a+c)ai—e, a+, a? — ae. 
GreC:) a—e, 1, a? — ae. 
1k ie a. 


Multiplying together a + c, a —c, and a, we find the 
least common multiple of the denominators to be a* — ae’. 
This is the least common denominator. 

The first denominator is contained in it a times; hence 
both terms of the first fraction must be multiplied by a. 


2 


Performing the operation, we obtain = 

The second denominator is contained in the common de- 

nominator a? —ac times. Multiplying both terms by this, 
en 

the second fraction becomes aoe 
a’ — ac 

The third denominator is contained in the common de- 

nominator a + ¢ times. Multiplying both terms by this, 


e+e 


; a 
the third fraction becomes = Rane Hence the fractions, 





reduced to others having the least common denominator, 
are, 


ELEMENTARY ALGEBRA, 129 


a? a’ — are ac + ¢ 
——_~ ~——— and ———.. 
a—ac a—ae’ a — ac’ 

EXAMPLES. 


Reduce the following fractions to others having the same 
values, and having the least possible common denominators. 
































a b ac b? 
1. b and ~ Ans. 7 and Bey 
Joys and g Atevene= ee ae bat) 
oid z LYZ UYzZ xyz 
4 
3. ee = and as Ans. tides ie and Le 
ax cx ac ace ace acx 
ax dx xy vx dx xy” 
A, Her ay’ and a Ans. ady’ ady’ an ady 
a+b a—b 
2 a+2 ps 
oo Gira —— be and — meee f Oe ale 
ag ee 
a+b a—b 
eee ss . 
Ang, CTL +B ae — Bab +B 
ns. Pane ae ey Te : 
ee a a + 53 a’? — P? a® +- 5° 
it e+ and a + b* Ans. ao +O and Pah BF 
8 we 2 5 ene 


ee vx ay —aby d a’x + abx 








a—b” a—bh’ Fed 
9. s and “4 Ans. a ae an oe. 
10. 2 S and = ye =e oe an ae 
ee car! aan sae ™ ate 


I 


130 


12. 


14, —— 


15. 


16. 


18. 


ELEMENTARY ALGEBRA, 


db ad Cx 


en ee “Or 9 2 an eo parts 
acx’y cba’y” axsy’ 














Pedy od 
ee wvbexy® a?bera*y” abexy® 
13 am em da 
‘ a'masy” acy” azxtm 
acm’ am*x*y edy® 
Ans. comizty® aemiaty? *"* emiaye 
Ta 9a 6a l4ax 9a 8 x? 
9 ’ Ao” and oye Ans. Awe Ag” and ae 
at+a Ax d 9a 
a—xata’ oe at 
i 2a7x? + 4ax’® + 2 xt 8 ax’ —8 a 
hee 2 a’? — 2 a 9 ax? —Qa% 
d 9a — 9 ax’ 

a Fiat — 2 aot 
ha C bdh adk bck 
iy B and i Ans. bak’ bdk’ and bdk 

eA Se Mn x 
EO ys RO erat 
Ane, 2%, 08 Bedm 5 4 aba 
' Pbed wbed wbed * wbed 
2 2 4 2 2 “| 
ees has and AS, Ang. oe hs and —. 
nn n n nn n n 
ax+b a b 
ME Pir at ones and oa 
a ax + b ac + ad be — bd 
me of —d? of —d?’ et e —d? 
s 2 2 
ae ab qv Ang owe b a 


20. 


BB wo ODS oe ab ab? ape 2nd oe 


ELEMENTARY ALGEBRA. ales f 


SECTION XXXV. 
Addition of Fractions. 


Frrsr reduce the fractions to their lowest terms, then to 
the least common denominator; add the numerators, and 
place the result over the common denominator. 

EXAMPLES. 


Find the values of the following expressions : 



































2 2 
zs - ~ Ans. 2 a 

2 2 4 

wyz2 ab’y  d’xyz xyz. 

8x 9a 6a 2427+ 9a _ 9a 
3. Or ae we Ans. Sey ee etry 6 + 4y° 
= ei ae Ans. 1 

ete 2 P -F-g 

a+b. a—b 2a? + 206? 
2 eg os An a? —}? 

a—sx@ a+z 2a 
6. = Pe + eae Ans Aas 

| 1 ee 1 2x 

Fe a + ey Ans. ey 


) a aa ap es me Ans. 0 
10 9a", 2abee | Tax 
“ Tabe 5a'b’xe  8ab'a 

) | vee 300 ba? + 112 be? + 245 a’e 





2 
Es edt + amx + ee a Eee Ans. 4ama. 


20. 


ELEMENTARY ALGEBRA. 























1 1 il 2a+1 
oe Rae a eT Ant 
ay aad thes ay® + y'2* + xia 
“2? se x? ap y” Ans. ea) fe 
"s iis a Ans. Ye eshte 
x y Zz LYz 
a+ b 4a b—a 4 a? 
AE Oo Ge Bl es Ans. Cu 
7 8 3 Tf+8g+3e 
—+—+ >. Ans. ~ v_>? 
og of | ig fy 
pole Di asy + bse + bry 
oF ak ta) iia Ans. bsy . 
a—b ¢#+0 a + b 38a + 3B 
Bers ah aie phan An. 
Cte a2 cat Zan ee 
ae ‘ata + 3 oe 

Aye eO+ar+art+est+2ad+ 22x 
8 rn 
EE Aiea) Ans. x. 





SECTION XXXVI. 


Subtraction of Fractions. 


PREPARE the fractions as in addition. Subtract the nu- 


merator of the subtrahend from the numerator of the minu- 
end. Set down the result over the common denominator. 





EXAMPLES. 
1. Subtract — _ from si , Ans. & 
2. Subtrace 18 ¢ from = a Ans 20 &. 


14 By “i ADT 


16. 


Li 


ELEMENTARY ALGEBRA. 133 









































rR a —b a+b 4ab_ 
. Subtract - + ; from aeocaegs Ans. my 
. Subtract u from : Ans. wT 
. Subtract Pe be from ; Ans. = 2y : 
ary ef x —y 
. Subtract — ab from — od Ans. 0. 
ab cd 
2 abex" pate a’bex 7 abema —4 abcx* 
9 mn 18 mv’ a 18 mn 
Subtract — ” fr rom a Ans wim age 
. Subtract > a Saat Mestre 
. Subtract SSE from ba 2 Ans. bce a 
oe? 2a a 
hee ene 2 
. Subtract Carte from eats: Ans. Was 4 poe 
a—Z ata a? — zx 
—b a+b 
ars from Mic Dab ake Ans. 0, 
1 1 a 
. Subtract ae from Ee Ans. — ce 
1 yi 1 
. Subtract a 6 from as Ans. pam Fea 
i af ec—b 
. Subtract “a from eer Ans. ico ghee 
2 1l+a 1 
. Subtract — TC. - from Eee, Ans. sparmyticsr pees 
2 — 
Subtract ~ from ea Ans. z rt! : 
z z Zz 


Subtract =e from laid Perms Yat ee ee 


12 


134 ELEMENTARY ALGEBRA. 











2 
18. Subtract sh sa from ——— Rating Chutes | Ans. 0. 
a—b a 4-6 | 
19. Subtract ” from ~. Ans. De ees ; 
J at xy 
a a a 
20. Subtract ore j from = Ans. Poe 
—0892.00——. 


SECTION XXXVII. 
Multiplication of Fractions. 


THE process is the same as in arithmetic. Multiply the 
numerators together for a new numerator, and the denomi- 
nators together for a new denominator, cancelling common 
factors of any numerator and any denominator. By such 
cancellation, the result will be in its lowest terms. 


EXAMPLES. 


Find the values of the following pee” 











1 18 a*bex cach . 25 aberx ve Be 
* 35 mn 27 ake’x 30 a®ca?" " 8a'nx 
12a°* 18 aby’ ab*c’y 
2 Tipe ae pee = mg 
2 
Bae tae Ans. = 
5 nae Y 
The Pie ac 
4, — b x A Ans ate es 
24c’m’n 121 ama 5 ya? veux 
: 132 aba’y ©" BS m 5 * {2 mn’ ae 6b- 
18 


6. —, of 5 atu. Ans. 90 a’ex. 
a’e 


ELEMENTARY ALGEBRA. 135 














3,2 
7. 11 amin?x x ASE 
mica 
eee 
; y’ mica “* 144 ate’x’ 
pee V1 ay? 
4) 51 ay! ene of — cx. 
31 ary 5b 1 
10. 21abe x 7 Be of 93ay x iF 
1 
11. —abe x ay xX any X 7. 
1g, Bam’ IL my’ 2 
* 15 ¢’my n> 302 na" 
13 Zor ye llutzy® | 24 cux 


Rie” 492° 


* "132 mée’q i 50 a*y <TR y’q- 


Cs} gptpn2 
21 c*m 18 y*c £91. 


da .a—2X .20bx 


‘3° ata. 3dabx 


ee). x pada x ( —d°). 


(et ay * =a) 














r+rs +3 2 et —s 
cS ,8 of 39 rs of 65 i 
a+b a—b : u 
Samp a bo oa 
ec—ax A Bie Cc—2x 
e—av°  e—2 +2? 
1 a —1 . 
pee yao 
cy x 
ce X cy ey 
Bede d)) 


Ans. 99 ac®n’y’. 











by 
Ans. — hot 
93 ax" 
Ans. — 3 yp 
3 
Ans. = 
Ans. — - 
am ny’ 
A 5 ean 
wryz 
Ans. — 18 mig? 
64973 
Ans = : oo 
ay 
15a—152x 
Ans. “a 
Ans. 1. 
38 
Ans. be 
Ans. a? —b? 
1 
ANS sae 
Ans. x’ 
Ans. ¢? 
Ans. y. 


136 ELEMENTARY ALGEBRA, 























98, oa of A of aoe Ans, 2% 
24. os r) oo of us + na wae ae 
25. nt of se of = Ans. ea = BRE. 
26. LS of ae oft . ae - Oe pe 
1 “of a5 > > se a’ : ab 
28, Coe of OT op OO’ Ans. 2. 
a9; SE + Bw Sat oe en 
30. — = ze aes snes ee _ 
SES ae 


SECTION XXXVIII. 
Division of Fractions. 


THE process is the same as in arithmetic. Invert the 
divisor and proceed as in multiplication. If the divisor is 
a compound fraction, invert every component fraction of it. 


EXAMPLES. 


Find the values of the following expressions: 


Say — 15 aPecy’ a4 
i Q2abea ~ Lata?’ AM, 6 a®y 
13 a’bed’ . oe 13 bd? 
2. OT aides. + 121a exy ° Ans. 2541 a abeny® 


Li. 


18. 


Ley 32 abex®y © 


ELEMENTARY ALGEBRA, 


25 a*bex® 

edz 
__ 33 minp”? 
125 aFyw 


+5 





—a--. 


a 


21 e’'dax® 





LL ated? | 
13 mny ° 


225 r'sdx . 





51 s*pq 
__ 31 mnop* 
‘5 abp’c 

18 ghk* 

5 g’e 

x — y? 
32+ 3y ° 
m? — n? 
Sat 
ee? + ed | 


of 18 = 





"ee — ed 
hay 
at bh 


am’r —amy . 


a+b 


a—b 





Oat 
12* 


cco 3 
Tabey © 
117 mcd 
99 na’ax 


ade, 150%8 | 
Silas 

Oe as 
Zab e 


_m—n 


“mtn 


abex. 


-- 121 mnpg. 


ees = hom ae 


2 ae'dy * 


3 ac’dxz? 
16 by’ 


— -- 3a'cd'. 


22 
3 cx” 


3 acmnop 


25 xty 





162. 


ity 





9 


e+d 


e—1 


ex? — cy? 
" vera + vey 


Ans. 


137 


Ans. 5a 


3 q° 
1375 xyu 
6 aetdx?y? 
. ne aes 


Ans. — 


Ans 





Ans. 1. 


cae 
" 3a'n?y 
eee 
" 34 pqrs® 
__ 31 boty 


25 acp 


2 hk? 
 Beg® 


Ans 


Ans 








a’ + 2ab + B? 


a — 2 aberan 
2 
Ans 





am 


on 


138 ELEMENTARY ALGEBRA. 





Ed aeons est COW 

19. Pe ” of eB See Eas SNP Ans. 1. 
ae +182+4+9 | Fre | | 

20. >a ea | Se 9a? —1824+9 Ans. 81x oh 81. 


SECTION XXXIX. 
Reduction of Complex Fractions to Simple Ones. 


THE process is the same as in arithmetic. First reduce 
mixed quantities to improper fractions, compound fractions 
to simple ones, etc., so as to have the numerator of the com- 
pound fraction nothing but one simple fraction, and the de- 
nominator nothing but one simple fraction. Then multiply 
the upper numerator by the lower denominator for a nu- 
merator; and multiply the upper denominator by the lower 
numerator for a denominator. The resulting fraction should 
be reduced to its lowest terms. 


a ne 


£4 





PROBLEM. Reduce to a simple fraction. 


ab 


a ne ne 
SoLurion. The numerator, — of ——, reduces to ——- 
C atx* aba* 





2 
The denominator, a + cass reduces to — ot a, The com- 
ab ab 
ne 
eee hott ie 
plex fraction then becomes —— Baha tetig 
ab 


We now multiply ne by ab for a numerator, producing 
abne. Then multiply a’b + 4c by act for a denominator, 


producing a®ba* + 4 acct. 


abne 


a’bxt + 4a3er* 


ELEMENTARY ALGEBRA. 


bne 


it to 


atba*t + 4 a’%ex*’ 
EXAMPLES. 


139 


Hence the resulting fraction is 


Both terms can be divided by a, reducing 


Reduce the following complex fractions to simple ones. 


Lg 


nS 


Sia} ore Slaloe 











x 
Ans. =: 


b 


140 


10. 


WE 


15. 


16. 


3 
Pay Ans. ke 
a 4a 
3 
1 
Hed 
3 Ans. = 
a 
amc Tm? 
ac'dx? Ans EE 
53 66 c'dx 
144 
PoE Ans iam 
— 1 
a 
2 
a 2 Ans, 000% 
eA Hie, 9987 
2mnr 
3 mv 
4 mn Ans, 1 
6 m’n 
1 
a Ans. = 
oe 
Fy Be oy 
a+ 2x a? — Ghee 
- 5 Ans. - 
i a+cx 
a + Ey 
a’ — o° 
1 a GE 
ey +a Ans. naan 
m—n 
of 
oe Ans. ds 


ELEMENTARY ALGEBRA. 
































7. 
dors atn 
a mn 


ELEMENTARY ALGEBRA. 141 














a 1 
c 18 @ aveta 
18. ‘ Ans. -—— 
a c 
= OL = 
Cc a 
Lae : ac + 
19. ss Ans Porat 
a+ - ‘ 
Cc 
1 
a a Ans ie 
7 i alah Nigk 
a 
—079400——. 


SECTION XL. 


Clearing an Equation of Fractions. 


If equals be multiplied by the same number, the results will 
be equal. 
Take 2 = 2. Multiply both of the equals by 3. Then 


we have 6 = 6. 
Take 5 = 3+ 2. Multiply both of the equals by 6. 


Then we have 30 = 18 + 12. 

Take} = 4+4. Multiply both of the equals by 4. 
Then we have 2 = 1+ 1. The equation has been cleared 
of fractions. — 

In order to clear an equation of fractions, multiply every 
term by the least common multiple of the denominators. 

: SP ice AE Sete ; 

PROBLEM 1. Given 7 dame tek Diet, + 42, to find the 
value of x. 

Sotution. The mixed number musi first be reduced to 


an improper fraction. Then we have - + emai: oth 


142 ELEMENTARY ALGEBRA, 


a The least common multiple of the denominators is 84. 
Multiply by 84. As it is a multiple of the denominators, 


every denominator may be cancelled in the multiplication, 


giving 21x + 28x —Tx = 12% + 360. 
Transposing, 
2lx + 28x2—T7a# —122 = 3860. 
30x = 3860. 
oii t AE 


: 1 Ses Os cael a 
PROBLEM 2. Given 9 ao ra +e rere: to find the value 


of x. 


SotutTion. Here, the least common multiple of the de- 
nominator is 36x. Multiplying every term by 36 x, we have 
4x+6x2-+ 9x = 86. 
19. v= 36: 
x= 11%. 
EXAMPLES. 


Find the value of x in the following equations: 


52 32 1 OO 
Loyal a> thig) da agen Ans. © = 24. 
ye 
2. 29% —28ax + Rar 8 —4}. Ans; a 
3: we 4440 = 88 —4e. Ans. « = 11. 
4, Sf 10 == —2 42 —32. Ans. « = 5. 
16 x 23 10x ; ie 
Dee rih chude Or ee Ans. x = 9. 
6. stl +2e=—5. Ans. « = —2J, 


1. E+ e=4a— >" + 80. Ans. 2 =e 


Nl 8 


ELEMENTARY ALGEBRA. 








x x 32 

ete 

A & 

g = 1 + 5. 

+25 = —15 

Tx a, “ : 

mee 8 Oe + 10g + 76}. 

Seerent. 993 

ee 5 ot 

zx+6 x—6 

2 5 ao 4 

ao oO = ae 

ae ae PL ie 
Va re ae 

te. We let 

5 55 2a. 

50 75 125 __ a, 

Bae Sx 8e 27 

‘i Se 

a+ 5 le 5 25 

2x2—T  38x2—2 

Beat a = x7—7 
Tx 5x (ms 
et 30 

Ee a ve a 

162 —68 47 


143 


ANS Meee: 
Ang oi 01, 
Ans. x = 40. 
Ans. x = — 80. 
ANS. el 
Ans, I= 24 
Ans. x = 86. 
Ans. & = 6, 
Ans. % cay). 
Ans, & = 2b, 
Ans. x = 61, 
Ans. x = — 358. 
Ang: Ciel in 
Ans. % = 93, 
Ans. x = — 16. 


Ans. x = 54. 


144 - ELEMENTARY ALGEBRA. 








24. 30” = 293. Ans. x = 138. 

25. een = wali saat Ans. 2 = 1 
2 3 

eo Deh alent ae, 

26. gy Aor aS ees od. Ans. e ==: 94, 

27. : 27 «2 + = = — 24,1). Ans. x = 33 

5x2 10x—21 mer 

28. EL ae — 82, Ans. x =, 82. 

29. 5a — 86 = « —171. Ans. & = 17%. 

x—8 Tax—A4 
30. ab er prt rel Ser et a ee Ans. x = 12. 
—059400——_ 


SECTION XLI. 


Questions producing Fractional Equations. 


PrRoBLEM 1. John has z as many marbles as George, 


when George gives John 36 marbles. Then 7, of John’s, 


plus z of George’s, amount to 50. How many had each at 
first ? | 


SoLution. Let 2 stand for the number of George’s mar- 
bles. 


Then : of x, or ake is the number of John’s. 


6 
When George gave John 36, George had left x — 36. 


John then had oe + 36. 


ELEMENTARY ALGEBRA. 145 


Now, we must find ‘ of John’s and . of George’s. 


7 5x : 3852 , 63 
ig f (| + 36) is aur 
2 A 2 x 
g of (x —36) is See rat 
Then we have the equation, 
Bie eed SS Ae 
Fe a a 


Clearing of fractions, 
105 x + 4536 + 64 « — 2304 = 14400. 
Transposing, 
105 + 64a = 14400 — 4536 + 2304. 
169 « = 12168. 


x= 72. 
Then ee = 60. Hence John had 60, and George had 72. 
1 h3 ae 
ProspitEM 2. At what o’clock does 9 of F of 7 of the 


time from noon equal the time to midnight? 


Sotution. Let x equal the time since noon, (in hours.) 


Then : of : of - of x is the time to midnight. 
te oe LD x 
Now, 5 of 5 of of x = 7 
Then, since the whole time from noon to midnight is 12 
hours, we have the equation, e+ 7 = 12. 
Clearing of fractions, Tx+ax= 84. 
8x2 = 84. 
x = 103. 


Hence it is half-past ten in the evening. 


18 K 


146 ELEMENTARY ALGEBRA. 


EXAMPLES. 


J, Find a number to which if its third part be added the 
sum will be 48. Ans. 36. 


2. Find a number such that if it is increased by 3 of it- 
self, 4 of the sum will be 22 less than the sum. Ans. 22. 


3. Find a number which being added to its fourth part, 
the sum will equal half the number, added to 15. Ans. 20. 


4, One-fifth of a cistern was filled with water. By add- 
ing 5 barrelfuls 1 of the cistern was filled. How many bar- 
relfuls of water would it contain in all? Ans. 100 barrel- 
fuls. 


5. A spent } of his money, and then received $1. He 
then spent 3 of what he had, and $8 remained. What had 
he at first? Ans. $20. 


6. Find a number whose half, third, fourth, and fifth 
parts added, are equal to 154. Ans. 120. 

7. Find a number whose sixth part exceeds its ninth part 
by 5. Ans. 90. 

8. B’s age is 211 times A’s. The sum of their ages is 47 
years. What is the age of each? Ans. A is 12 years old, 
and B 35. 

9. Divide $60 among A, B, and C, giving B 3 as much 
as C,and A 2.as much as C. Ans. A must receive $12; B, 
$18; and C, $30. 

10. Divide $60 among A, B, and C, giving B 3 as much 
as C,and A 4 as much as B. Ans. A must Ben $10; B, 
$20; and C, $380. 


11. A merchant lost 4 of his capital during the first year. 
The second year he gained 2 as much as he had left at the 
end of the first. The third year he gained ;3 of what he 
had at the close of the second, making his cannes $7000. 
What was it at first? Ans. $5000. 


NotE. Let z = his capital at first. Pe lost ~ and had te re- 


ELEMENTARY ALGEBRA. 147 


4 f 4x 
maining at the end of the first year. Get 3 of ee and add it to —, 
5 5 


to get what he had at the end of the second year. Get then 5% of 
that result, and add it to that result to get what he had at the end 
of the third year. 


12. My capital is $2220, being 11 per cent. greater than 
it was last year. What was it last year? Ans. $2000. 
Nore. 11 per cent. is +4. 


13. Divide 32 into two parts such that 1 of one may equal 
4 of the other. Ans. 12 and 20. 


14, Find a number whose fourth plus its third less 12, 
equals its half less 6. Ans. 72. 

15. At what hour does } + 1 + 3 + ,5 of the time past 
noon equal the time to pass before midnight? Ans. 4 P.M. 


16. Divide 325 into two parts, one of which is 12 times 
the other. Ans. 125 and 200. 


17. If you multiply a certain number by 3, add 4 to the 
product, divide the sum by 5, and subtract 7 from the quo- 
tient, the remainder will be 10. What is the number? 
Ans. 27. 


18. If you add 6 to a certain number, divide the sum by 
9, add 16 to the quotient, and multiply the sum by 3, the 
product will give the number itself. What is it? Ans. 75. 


19. A’s capital was at first 3 of B’s. A gained $100 and 
B lost $100. After this 14 of A’s capital added to the 
whole of B’s amounted to $6500. What had each at first? 
Ans. A had $3300, and B $4400. 

20. A man spent $14 more than 3 of his money, and had 
$6 more than $ of it left. What had he at first? Ans. $84, 

21. A and B have the same income. A saves 10 per 
cent. of his. B spends $75 a year more than A, and saves 
$125 in 5 years. What is the income? Ans. $1000. | 

22. A merchant sold his stock, good-will, and fixtures for 
$50,000, receiving 5 times as much for the good-will as for 


148 ELEMENTARY ALGEBRA, 


the fixtures, and 34 times as much for the stock as for the 
good-will. What did he receive for each? Ans. $38,000 
for the stock, $10,000 for the good-will, and $2000 for the 


fixtures. 


23. From a certain sum I took its fourth part, and then 
added to the remainder $29. I then took away the tenth 
part of this, and afterward added to the remainder $25, 
which made it $70. What was the original sum? Ans. 
$28. 


24. A young man spent } of his yearly income for board 
and lodging, 4 of the remainder for clothes, and 4 of what 
was left for amusements. The rest, $400 per annum, he 
saved. What was hisincome? Ans. $1600. 


25. A laborer was engaged for a year at $440 and a suit 
of clothes. He left at the end of 5 months, receiving $160 
and the suit of clothes. What was the value of the suit of 
clothes? Ans. $40. 


Nore. What he received was ;; of a year’s wages. 


26. A bought stocks amounting to $3000. B bought 
stocks amounting to $2000. A sold a certain amount of 
them, and B sold half as much as A. A then had 1} times 
as much as B. How much did each sell? Ans. A sold 
$1500 worth, and B, $750 worth. 


27. A quantity of pure water being decomposed, 3 of it 
by weight plus 125 ounces was oxygen, and ,'; of it by 
weight less 15 ounces was hydrogen. What was the quan- 
tity of water? Ans. 90 ounces. 


Note. If 2 represent the quantity of water, the two parts of it 
added together equal z. 


28. A father divided a certain sum of money among his 
four children at a fair. To the oldest he gave 65 cents less 
than } of it; to the second, 25 cents less than 4 of it; to 
the*third, 9 cents more than | of it; and to the fourth, the 
remainder, which was 70 cents. What was the whole 
amount? Ans. $3.30. 


ELEMENTARY ALGEBRA, 149 


29. Divide the number 22 into four parts such that if the 
first be increased by 4, the second multiplied by 4, the third 
diminished by 2, and the fourth increased by 2, the results 
shall be equal. Ans. 4, 2, 10, and 6. 


Notre. Let x stand for any one of the equal results. Then x —4, 


x + 2, and x —2 are the parts. 


30. In a certain quantity of Chinese gunpowder, the nitre 
was 19 lbs. more than } of the whole, the charcoal was 163 
lbs. more than 3 of the whole, and the sulphur was 29 lbs. 
less than 1 of the whole. What was the amount of gun- 
powder? Ans. 162 lbs. 

31. Divide $870 between two men, so that 34 of what the 
first receives will be equal to = of what the second receives. 
Ans. The first receives $370; the second, $500. 

32, A merchant buying goods spent at one store } of his 
money and $300 more, when he had $3000 left. How much 
had he at first? Ans. $4400. 


33. Divide 63 into two such parts that if the less be di- 
vided by 5 and the greater by 4, the sum of the quotients 
will be 15. Ans. 15 and 48. 


34, An estate is to be divided among three children so 
that the first shall have $1400 less than half its value; the 
second, $1300 more than } of its value; and the third, 
$840 more than } of its value. What isits value? Ans. 
$14,800. 

35. A man has a lease for 39 years, and 2 of the part of 
it which has expired is equal to .5, of the part yet to run. 
How long before it expires? Ans. 24 years. 

36. What number is that whose } added to its 1 makes 
382? Ans. 60. 

37. A, B, and C gave a certain sum of money to the 
poor. A gave i of it; B,4; and C, $50. What wasgthe 
sum? Ans. $120. 

38. A lived in his house 24 years, during which time an 

13 * 


150 ELEMENTARY ALGEBRA. 


addition was made to it. One-third of the time he lived in 
it before the addition was made, is equal to 4 of the time he 
lived in it afterward. How long is the latter time? Ans. 
15 years. 

39. After paying away 4 of my money and § of what re- 
mained, I had still $75. How much had I at first? Ans. 
$120. 

40. What number is that from which if 21 be subtracted 
3 of the remainder will be 60? Ans, 121. 


41. After paying away } and 3 of my money, I had $84 
left. How much had [ at first? Ans. $144. 


42. A gamester staked 3 of his money, which he lost, but 
afterward won $3, when he had $42. How much had he 
at first? Ans. $52. 


43. A gentleman spent in one year 2 of his income for 
the support of himself and his family, and 2 of the remain- 
der for a piano. ‘There was $200 left. What was his in- 
come? Ans. $1500. 


44. A horse and a chaise are together worth $800, and 
the chaise is worth 13 as much as the horse. What is the 
value of each? Ans. The chaise is worth $509, and the 
horse, $300. 


45. A house rents for $924, which is 10 per cent. more 
than last year. What was it last year? Ans. $840. 


46. A and B have the same income. A contracts an 
annual debt amounting to 3 of it. B lives on 3 of it. At 
the end of 3 years B lends A enough to pay his debts con- 
tracted in that time, and has $135 of his savings left. What 
is the income of each? Ans. $540. 


Nore. Interest is not to be counted either on debt or savings. 


47. What number is it whose 1, 3, and 2 are together 
equal to 92? Ans. 105. 


48. A person, after spending $300 more than 4 of his in- 


ELEMENTARY ALGEBRA. 151 


come, had remaining $200 more than $ of it. What was 
his income? Ans. $900. 

49, Find a number such that if it be increased by 9, the 
sum shall equal ? of the number plus 13. Ans. 16. 

50. From a purse containing a certain sum there was 
taken $10 more than its half, and from the remainder, $3 
less than its fifth part, when $55 were left. How much did 
it contain at first? Ans. $150. 


SECTION XLII. 


Review. 
EXAMPLES. 
1. Divide 21 ay by 35 a*y'n. Ate oe 
2. Add together abe, acb, and cba. Ans. 3 abe. 


3. What are the factors of a? — <x’? 
Ans. a + canda—ze. 
4, What are the factors of 25 a?b"c* — 9 d®etf?? » 
x Ans. 5 a’b’ct + 3 d’e?f and 5 a®bict — 3 d¥e’f. 
5. What are the factors of 4y’ + 1242? + 92*? 
Ans. 2y+ 382 and 2y+ 32% 
6. What are the factors of n®y? — 30 mintry + 225 mx?? 
Ans. nty —15m'’x and nty —15 mix. 
7. From —8 2 —17 ay’? + 5 take — 82° + 5 ax’y — 3. 
Ans. —17 xy? —5a’y + 8. 





8. Add —a.to —b. Ans. —a—Ob. 
n? 
9. Reduce : aie to a simple fraction. Ans. _ 
. a 


a + ab 


152 


10. 


AL. 


12. 


13. 


14, 


15. 


ELEMENTARY ALGEBRA. 


xa+27  «—138 x—23 . 94+ 67 























Given, “907+ 20. = 4b ot to 
find the value of «. Ans. © = —Y7, 
Divide 12 a’c’x* — 16 a%e’x? + 9 abc’x* by —2 atex*. 
Ans. — 6 abex + 8 aca? — ? —_ 
he 
= 4 of —--- to a simple fraction. 
Ans. ae 
yz 
FN fi AA SEAN ld Ans. 0 
mn mp np 
2 Le eee 
Find the value of x in the equation > tee 6. 
Ans, © =e 
What is the value of the expression 3 + 6 x 9 —5? 
Ans. 52. 
. What is the value of the expression (81 — 4) + (4 + 
3 xX 9 —A4)? Ans. 1. 
. Multiply 6 —1 by 6 — 2, and subtract 5 from the pro- 
duct. Ans. b?’ —4b + 2. 
| What is the value of © — 72 4 % _ — ee 
ax cr 6, am 
3T 4 Ans. 4 
7b 
9,2 8 aw? 
. Reduce oe 4 ats to its lowest terms. Ans. aa 
. Divide a® + aa + ax” + x¥ by a? + 2”. 
Ans. a + a. 
a+b @ eed a” — ? a? = ae | 
. Multiply together — i pt AL Gab ee and 
ee 2 _ 
a 2ab + 6 Ase a —b 








Be ka pias 4 


ELEMENTARY ALGEBRA, 153 





22. Find the value of x when x aioe = (0, 
Ans. x = 5. 
23. Find the value of (9 abe —16y’ + 3rs) —(12abe + 
5y? —rs + abc). Ans. —4abe —21y’? + 4rs. 
.., a —F a—bd a+ b 
24, Divide TTS by ciiaiie Ans. aM: 


29. 


30. 


. What is the number whose fifth part divided by 3 is 


equal to 7? Ans. 105. 


. Divide a’ + 2acx + ex’ by a + ex. Ans. a + cx. 
. If 40a” is equal to 16 x (2x + 1), what is the value 


of x? Ans. x = 2, 


. Reduce x — y — — to the form of a fraction. 


Py sma Y yeas til 
yp ities kad Aik Brac 





a -2 Y 
x? 2 z 3 F 
Reduce Se to its lowest terms. 
1 
Ans. : 
a—y 


Add 5 m'n'p + 3 m’n'p? — 11 m‘ndp, 3 m'n'p + min'p 
— dm’n*p’, 3 min*p + 16 m?n3p? — 12 m‘nip, and 
3 mn'p*? + 21 min'p — 5 min'p. 

Ans. 32 m’n'p + 17 m’n*p? — 27 m‘n'p. 


REVIEW QUESTIONS. 


What is an exponent ? a coefficient? Name the exponents in the 
expression 24 a8i?74. What do they show? 

What is a monomial? a binomial? a trinomial? a polynomial? Is 
a binomial a polynomial ? 

Write two monomials, two binomials. 

Isa+<«capolynomial? Is 36 az? a polynomial? 

In what ways is multiplication expressed ? 

In what ways is division expressed ? 

Explain the use of a parenthesis or of a vinculum. 


What is 4/25? /xty?? W238? 


154 ELEMENTARY ALGEBRA. 


What are similar terms? How are they added? 

Explain the process of subtracting — a from e¢. 

What is the reason that a quantity which is transposed from one 
side of an equation to another, must have its sign changed ? 

Explain the multiplication of — 4 by 6. 

What do we mean by multiplying by a negative quantity ? 

What is the rule for signs in multiplication and division? 

What is the rule for indices in multiplication? in division ? 

Explain the division of —z by —y. 

What is meant by elimination? When is it necessary ? 

What is the square of the sum of two quantities equal to? the 
square of the difference? the product of the sum and the differ- 
ence? 

What is meant by factoring? What is a measure of a quantity ? 
What is a prime quantity ? 

What is a multiple of a quantity ? 


SHOLION elas: 
Elimination. 


Ir has been already explained that when we have two 
equations, containing two unknown quantities, one of them 
must be eliminated before the equations can be solved. Three 
methods of doing this are in common use. One of them, 
elimination by addition or subtraction, has been already ex- 
plained. They will, however, all be brought together pee 
for the sake of a comprehensive view. 


EXAMPLE. Given 
ie ay eteg 1 5x 
S.A eel 
(x —10) (y —6) = 5 + (& —5) (y —8) 
to find the values of 7 and y. 


The equations must first be brought into the form proper 
for elimination. 


ELEMENTARY ALGEBRA. 155 


Clearing the first equation of fractions, we have, 
48a —45 = 8y —72 + 452. 
Transposing, 
482 —45x —8y = —72 + 46. 
38a —8y = —27. 


Performing the operations of multiplication indicated in 
the second equation, zy —6x —10y + 60 = 5+ czy 
—8x2—5y + 40. 

Cancelling xy, and transposing, 
—6x2—10y 4-8x+ 5y = 5 + 40 — 60. 
2% —d5y = —15. 

The equations are now reduced to forms suitable for 
elimination ; all the unknown quantities are on the left of 
the sign =, and they are collected into as few terms as 
possible. This must be done before any method of elimination 
is used. We will now take these equations and show how 
the values of x and y may be found in three different ways. 


1, Elimination by Addition or Subtraction. 


382—8y = — 27 
2x—5y = — 15 


Multiplying the mem- 


bers of the first by 2, 62—16y = — 04. 
Multiplying the mem- 

bers of the second by 3, 62—1dy = — 45. 
Subtracting, —y = — 9. 
Or, Gist o, 

Substituting —5 times 9 for —5¥ in the second equation, 
2x2 —45 = — 15. 
Transposing, 2x2 = —15 + 45. 
ESN U3 


Hence x = 15and y = 9. 


156 ELEMENTARY ALGEBRA. 


2. Elimination by Substitution. 
8a—8y = —27. 
2x2—dy = —16. 

In this method, find the value of one of the unknown 
quantities in one equation, just as if the other quantity were 
known. Thus in the above equations, 2 2 being the simplest 
term, find the value of x in the equation that contains 2 x. 





Thus, 2x —dy = —10. 
Transposing, 22 = Sy — 165. 
teas 5y —15 
Dividing by 2, 2 ea 
Now the other equation contains 3x. Multiplying the value 
of x just found by 3, we have of = em 
Substituting this instead of 32 in the first equation, we 
have irae —8y = — 27. 


This equation contains but one unknown quantity, and 
can be easily solved. 


Clearing of fractions, 





15 y —45 —16y = — 54. 
Transposing, 15y —16y = — 54 + 46. 
Collecting, —you PD, 
Whence, ys: 
We found before that C= oY oe : 


Putting 45 for 5 y, we have 
45—15 30 


Hence x = 15 and y = 9, as before. 


This method is generally longer than the first, though 
when the coefficient of x is unity, and in a few other cases, 
it may be the shortest. 


ELEMENTARY ALGEBRA. 157 


3. Elimination by Comparison. 


BS nee, 
Diese Bafa 15; 


In this method find the value of the same unknown quan- 


tity in both equations, just as if the other quantity were known. 
Thus, 


32 —8y = — 27. 2x2 —b5y = —15. 
3a = 8y —27. 2x = 5y —15. 
eS ee 0 ure ty 
anaes aes sss TS A 


Now, we have here two values of x; and as they are 
values of the same thing they must be equal. Hence we 


— 5y —15 
have a new equation, SB Beg CLIN Ire 2s 





3 2 
Clearing of fractions, 16 y —54 = 15y — 45. 
Transposing, 16 y —1dy = 54 — 45. 

Yy = 
Now, as before, 7 = ee 
Putting 45 for 5 y, we have 
45—15 30 
Se a 15. 


Hence x = 15 and y = 9, as before. 


Rule for Elimination by Addition or Subtraction. 


Multiply or divide the members of each equation by such 
numbers as will make one of the unknown quantities have 
equal coefficients. 

Then add equals to equals, or subtract equals from equals, 


so as to cancel or eliminate that unknown quantity. 
14 


158 ELEMENTARY ALGEBRA. 


Rule for Elimination by Substitution. 


Find the value of one of the unknown quantities in one 
of the equations. 
Then substitute it for the unknown quantity in the other. 


fule for Elimination by Comparison. 


Find the value of the same unknown quantity in both 
equations. 
Then place these two values equal to each other. 


EXAMPLES. 


Solve the first five examples by each of the three methods. 
1. Find the values of and y, when y+ wv = 6 and 


y—x = 4, Ans, 2 =e 
9. Find the values of x and y, when 4a + 2y = 20 and 
3”2—dy = —l11. Ans. 2 == 5 5 nee ae 
3. Given 139 ENP z a at to find the values of y and z. 


Ans. y = 13 fee 


4, Find the values of x and y in the following equations: 
eee eee Ans. © = 8; y= 2. 


3 \ to find what x and y are 
equal to. Ans. © = 133 ¥ =a: 


In the following pairs of equations find the values of x 
and y by whatever method seems most convenient. 


Ei eaanannoca set Ans. «= 6; y =2. 
4 2A == 2. 

if Eyairsomen ‘ Ans. 2 Sy ae 

8. sia aa eal Ans. c= 6; y = —6. 


10. 


12. 


13. 


14. 


15. 


16. 


ELEMENTARY ALGEBRA. 159 


Gay — 244 af . 
2 Ceo ; Ans. © = 4; y= 6. 
by —3x = 18. 
oui ay 13 y — 80 
oe gee 
ety a ee 
PRT : 


Ans. Gi 9: y = 16; 








Bet 4 Leys 
esis Sen 
| Ans. © = —2;y=1 
Sa+4  38y—a2 | Ty —62—5 
[oars Ce Bes CE 
oo TY 39 —y—15 
Ans 2=7;, y = —3 
2xa—3y 8x—D9y 
es oa 
Se ae 7 
1 me ane — 
Soegiciaek Shae 
Ans, x=, 44; y= 


Gat+4y 2_1 
FE Ree 
fe 175 = Oy + 62. 


Ans. x = 174; y = —17}. 





LEU —38y+4e—11. 
252+ 2y 
Pho ae 
Ans. = 43.4 == 2. 
llxw+9y 9 
99 aie 
Soe +10y=1lxe + 12. 


ANS. (0 ==. Ois yee 1 T. 


160 ELEMENTARY ALGEBRA. 





17. 








8a2—8 
ob 
ety = 34. 


ee eee ee 
18. Pred be 


Ans. 0.%=\21 4) ee te, 


19. 





20. 











22. 


23. 


24. 


25. 


| 
Han “i 
i ies tn >< 


2x—dsy= 0.| Ans. © = 8; y = 2. 


ELEMENTARY ALGEBRA. 161 


SECTION XLIV. 


Questions producing Equations with Two 
Unknown Quantities. 


1. Fryp two numbers such that { of the first plus 3 of 
the second equals 9, and } of the first plus 1 of the second 
equals 8. Ans. 20 and 15. 


2. Find two numbers such that the greater plus } of the 
less equals 23, while the less minus | of the greater equals 
11. Ans. 20 and 15. 


3. Find two numbers such that 5 of the first less 3 of the 
second equals 8, and 4 of the first less ; of the second 
equals 48. Ans. 720 and 512. 


4, There is a number represented by 2 digits whose sum 
is 9; and double the number added to 18 gives another 
number represented by the same digits in an inverted order. 
What is the first number? Ans. 27. 


Notr. See the note to Example 4 of Section 29. 


5. A man has two horses; also a saddle worth $50. The 
value of the best horse and the saddle is double the value 
of the other horse. The value of the latter horse and the 
saddle is 4 of the value of the first horse. Required the 
value of each. Ans. The first horse is worth $250; and the 
other, $150. 

6. Two casks contain wine; and 3 of the contents of the 
first is 20 gallons less than § of the contents of the second, 
while 4 of the contents of the first is 2 of the contents of the 
second. How much does each contain? Ans. The first con- 
tains 84 gallons; and the second, 63 gallons. 

7. Find a fraction whose value is 7 if 19 be added to its 
numerator, and is ;1 if 19 be added to its denominator. 
Ans. 2. 

Note. Represent the fraction by 7 


14* L 


162 ELEMENTARY ALGEBRA. 


8. Find a fraction such that 5 being subtracted from both 
terms, its value will be 4; and 3 being added to both terms, 


its value will be ;43. Ans. 3}. 


9. A and B together have $12,000. If A lose 4 of his 
capital and B gain } of his, they will have the same. How 
much has each? Ans. A has $7000, and B has $5000. 


10. A boy bought 46 cents’ worth of balls and tops, giv- 
ing 4 ee for each ball and 2 cents for each top. He then 
sold =; of the tops and } of the balls for 20 cents, clearing 
6 or on them. How many of each did he buy? Ans. 
He bought 6 balls and 11 tops. 


Nore. Let x= the number of balls, and y the number of tops 
4 2 

which he bought. Then4z-+ 2y = 46 cents. Also, > a aie 

what he gave for those which he sold. But he cleared 6 cents on 


+ 2 
these. Hence, ze of =+ + 6 = 20. 


11. A grocer bought $1.23 worth of watermelons and 
cantaloupes, at the rates of 12 cents and 3 cents. He then 
sold + of the watermelons and 4 of the cantaloupes for 50 
cents, clearing 17 cents on them. How many of each did 
he buy? Ans. He bought 8 watermelons and 9 cantaloupes. 


12. A owes $700, and B owes $600. A’s money plus 3 
of B’s would pay A’s debt. B’s money plus 7% of A’s would 
pay B’s debt. How much money has each? Ans. A has 
$650, and B has $250. 


13. A merchant bought two casks of wine, the first at 
$1.25 and the second at $1.50 a gallon, the whole costing 
$147. He sold 4 of the first and § of the second for $140, 
gaining $29 on what he sold. How many gallons were in 
each cask? Ans. There were 42 gallons in the first, and 
63 gallons in the second. 


14. Five years ago, A’s age was 21 times B’s. One year 
hence it will be 14 times B’s. How old is each now? Ans. 
A is 12 years old; B, 8. 


“ELEMENTARY ALGEBRA, 163 


15. Find a fraction such that if the numerator be dimin- 
ished by 5 and the denominator by the numerator, its value 
will be 1; but if the numerator be increased by the denom- 
inator and the denominator by 33, its value willbe 4. Ans. 
ote 
i fg 

16. If A should give to B as many books as B has, and 
then B should give A + as many as A had remaining, each 
would have 250. How many has each? Ans. A has 350 
books, and B. has 150. 


17. Divide $100 between A and B, giving B 3 times as 
much as A. Ans. A must have $25; and B, $75. 


18. Find two numbers whose difference is 6; and if 4 of 
the less be added to 4 of the greater, the sum will be equal 
to 4 of the greater diminished by 4 of the less. Ans. 18 
and 24, 


19. A man bought a house and lot for $3500. Two- 
thirds of the price of the lot was equal to 4 of the price of 
the house. What was the price of each? Ans. The lot 
cost $1500; and the house, $2000. 


20. Divide 189 into two parts so that 4 of the greater 
minus 3 of the less shall be equal to ;% of the less minus 2 
of the greater. Ans. 84 and 105. 


21. A man bought two pianos for $880. If of the price 
of the first be subtracted from 4 of the price of the second, 
the remainder will be 2 of the difference of the prices. 
What was the price of each? Ans. The first was bought 
for $540; and the second for $340. 


22. What fraction will equal 4 if 2 be added to its nu- 
merator, and 4 if 12 be added to its denominator. Ans. 4§, 


23. A certain number consisting of two digits, on being 
divided by 5 gives a quotient equal to 4 times the unit digit, 
and a remainder of 3. The same number on being divided 
by 18 gives a quotient equal to half the digit in the tens’ 
place, and a remainder of 9. What isthe number? Ans. 
63. 


164 ELEMENTARY ALGEBRA. 


24. If A were to give B $300, B would have 3 times as 
much as A has left. If B were to give A $300, A would 
have 3 as much as B has left. How much has each? Ans. 
A has $700; and B, $900. 


25. Two brothers had just sufficient money to purchase 
a certain house. The elder could purchase it with his own 
money, 3 of his brother’s, and $1000. The younger could 
purchase it with his own money, i of his brother’s, and 
$2000. How much had each? Ans. The elder had $4000; 
the younger, $3000. 


SECTION XLV. 


Equations containing more than Two Unknown 


Quantities. 
x+t2Qy+32z = 382 
PrRoBLEM. Given {22+ y+4z = 37> to find 
3x+5y+2z2 = 49 


the values of x, y, and z. 


SoLutTion. Here there are three unknown quantities. 
If we eliminate one of them, we will have only two. But 
when we have two unknown quantities, we must have two 
different equations. To get these we will have to perform 
two processes of elimination. Here the coefficients of x are 
smallest, and such as to make it best to eliminate x. This 
will be best done, first, from the first and second ete 
and then from the first and third equations. 


Multiplying the members of the first equation by 2, 
2x+4y+ 6z= 64. 
The second equation is, 2% + y+ 4z= 87. 


Subtracting, 3y +2z= 27. 








ELEMENTARY ALGEBRA. 165 


Multiplying the members of the first equation by 3, 
32+ 6y+ 9z= 96. 
The third equation is, 8x2+ 5y+2z = 49. 
Subtracting, . yt+tz= 47. 
By these two eliminations of x, we have obtained two 
equations with two unknown quantities: 


oy +22 = 27. 
y tz = 47. 


These can be solved as has been previously done. 


Multiplying the members of the last equation by 3, 
The equation preceding it is 3y +22 = 27. 








Subtracting, LO 2c: 
z= 6 
Substituting 2 times 6 for 22, we have 
3y + 12 = 27. 
3y = 27 —12, 
dy = 15. 
y = 5. 


In the first equation, substituting the values of y and z, 
x + 10+ 18 = 82. 
x = 32 —10 — 18. 
© = 4, 


Therefore « = 4, y = 5, and z = 6. 


General Rule for Elimination. 


Eliminate one of the unknown quantities from as many 
different pairs of the given equations as there are other un- 
known quantities. 

Do the same with the new equations obtained, till finally 
but one unknown quantity will remain in one equation. 

Find the value of this, and then of the others by substi- 
tution in previous equations. 


Nore. There must be as many equations as there are unknown 
quantities, and each equation must be used in one of the elimina- 
tions. The equations must be independent. See page 90. 


166 ELEMENTARY ALGEBRA. 


EXAMPLES. 


Find the values of the unknown quantities in the follow- 
ing equations. 


e+ y+2z= 115. 

1. xrt2Qy— z= 70. 
=e, 

Ans. % = 35; y = aUsieenee 


Ans. = 20; y= 8; 2=3. 
Ta—Qy+2z= 23. 
4, x— yt+z= 12. 
Qea+ yt2z2= 29. 
Ans. 7 = 3; y = 77 


= 21. 
x—9y + 8z= — 6. 
—— 6. 
Ans. 2 = 383; y= 7; 2=6. 


ce hae or 
ctiaya eens 11, 
Sat a ee 
6. Ania 9. 
a +f +2 = 183 
Ans. & = 12; 4 =e 
Y : 
TROD oh 259 y iA 
TR TF oe 4. 
L+y tz = 39. 


Ans. x's 24 34) =O eee 


ELEMENTARY ALGEBRA. 167 











PiU es ihm p Ati | hl Mie ae — tae PE 


ety + z=2137. 
11. 4¢a—ytz= 7. 
Sty t= 145 


Ans. = 73; y= 15; 2 = GH. 


Norse. In this example, add the members of the second equation 
to those of the third, and find z; subtract the members of the sec- 
ond from those of the first, and find y; subtract the members of the 
third from those of the first, and find z. 


et yt4z= 12. 

2} xt Qy+ = 

2e+ 8y4+42= 24. 
is Cee 4 of ae Aaa 


Yy Zz 
13.4 2+2y+2z=10 
gtyti2z= 9 


AR ee et) coed a cee, 


168 ELEMENTARY ALGEBRA. 


sy+2z= 28. 

Ans. 2 = 5; yo le Zee 
( elytra alt 
15. 4—ax2—ytz=—1. 
—a+y—z=—3. 


Ans. 0 == 2: eee z= 4, 
Notr. This example may be solved in a similar manner to that 
pointed out for the 11th. 


SECTION XLVI. 


Questions producing Equations with more than 
Two Unknown Quantities. 


1. A Boucut three houses. The price of the first plus } 
the price of the other two was $2200. The price of the 
second plus 4 of the price of the other two was $2300. The 
price of the third plus 3 of the price of the other two was 
$3000. What did he pay for each? Ans. He paid $1100 
for the first, $1900 for the second, and $2500 for the third. 


2. A and B together have 2 as much money as C. A and 
C together have 43 times as much as B. A has $6600 less 
than B and C together. How much has each? Ans. A 
has $900; B, $1500; and C, $6000. 

3. A’s money exceeds ,;% of B’s and C’s by $100. B’s 
money exceeds 4 of A’s and C’s by $100. C’s money ex- 
ceeds ;% of A’s and B’s by $100. How much has each? 
Ans. A has $800; B, $900; and C, $1000. 

4, A number expressed by three digits, gives 96 for a 
quotient when divided by the sum of the digits minus 9. 


ELEMENTARY ALGEBRA. 169 


The middle digit is half the sum of the first and third. If 
396 be subtracted from the number, the remainder will 
consist of the same digits as the first, but in an inverted 
order. What is the number? Ans. 864. 


Note. Let the digits, in order, be represented by 2, y, and z. 
Then 100 z + 10y + z represents the number. 


5. A number expressed by three digits is 22 times their 
sum. If 99 be added to it the result will consist of the 
same digits as the number, but in an inverted order. The 
middle digit is equal to the sum of the first and third. 
What is the number? Ans. 132. 


6. Find four numbers such that once the first, } of the 
second, 3 of the third, and } of the fourth shall be 23; twice 
the first, once the second, 3 of the third, and 3 of the fourth 
shall be 41; three times the first, twice the second, once the 
third, and three times the fourth shall be 111; and the sum 
_ of the numbers shall be 51. Ans. 8, 12, 15, and 16. 


7. Find three numbers such that the first with ;3, of the 
difference of the other two shall be 42; the second with ¥, 
of the sum of the other two shall be 120; and the sum of 
the three shall be 204. The first number is the smallest, 
and the third is the largest. Ans. 24, 60, and 120. 


8. A certain sum of money was divided among A, B, and 
C, so that A’s share exceeded ;4, of the sum of B’s and C’s 
shares by $200; B’s share exceeded # of the sum of A’s and 
C’s shares by $300; and C’s share exceeded ;, of the sum 
of A’s and B’s shares by $400. What was the share of 
each? Ans. A’s share was $600; B’s, $500; and C’s, $800. 


9. If A’s age be added to B’s, the sum will be C’s age. 
If A’s be subtracted from B’s, the remainder will be 3; of 
C’s age. The sum of their ages is 140 years. How old is 
each? Ans. A’s age is 34 years; B’s, 36 years; and C’s, 
70 years. 

10. A’s money added to ,*, of the sum of B’s and C’s 

15 


170 ELEMENTARY ALGEBRA. 


equals $55,000. B’s added to ;8 of the sum of A’s and C’s 
equals $80,000. C’s added to ;3; of the sum of A’s and 
B’s equals $60,000. How much has each? Ans. A has 
$30,000, B has $40,000, and C has $45,000. 


11. A certain sum of money is to be divided among A, 
B,andC. A is to receive $1000 less than 4 of it; B, $1000 
more than } of it; and C, $1000 less than 4 of it. What is 
the sum, and what is each to receive? Ans. A is to receive 
$5000 ;- B, $4000 ; and C, $3000: in all, $12,000. 

12. A farmer has sheep in three pastures. The number 
in the first, added to 4 of the number in the other two, 
makes 46. The number in the second, added to 4 of the 
number in the other two, makes 40. The number in the 
third, added to ;4; of the number in the other two, makes 
30. How many are in each pasture? Ans. There are 35 
sheep in the first, 30 in the second, and 25 in the third. 

18. A certain loaf of bread is made of flour, rice, and 
water, and weighs 11 lbs. The weight of the rice augmented 
by 1 lb. is + of the weight of the flour and water together, 
and the weight of the water is 3 of the weight of the flour 
and rice together. Required the weight of each. Ans. The 
rice weighs 1 lb.; the flour, 5 lbs.; and the water, 5 lbs. 

14, There is a number consisting of three digits, the first 
of which is } of the sum of the other two. If 99 be added 
to the number the digits will be inverted. The middle digit 
is 1 more than the sum of the other two. What is the num- 
ber? Ans. 142. 


15. A merchant sells 2 lbs. of his best tea, 5 lbs. of his 
best coffee, and 7 lbs. of his best sugar for $7.40. He sells 1 
lb. of the tea, 2 lbs. of the coffee, and 5 lbs. of the sugar for 
$3.80. He sells 3 lbs. of the tea, 2 lbs. of the coffee, and 4 
lbs. of the sugar for $7.60. What is the price of each arti- 
cle? Ans. The tea is $2 per lb.; the coffee, 40 cents; and 
the sugar, 20 cents. : 


16. Divide 50 into three such parts that 3 times the first 


ELEMENTARY ALGEBRA, bil 


shall be twice the sum of the second and third; and the 
second plus 14 shall be equal to the sum of the first and 
third. Ans. 20, 18, and 12. 

17. A’s age is twice the sum of his son’s and his daugh- 
ter’s. The three ages amount to 60 years; and 3 times the 
daughter’s plus 4 years is equal to the difference between 
her father’s and her brother’s age. What is the age of 
each? Ans. A’s age is 40 years; his son’s, 12 years; and 
his daughter’s, 8 years. 

18. A drover has oxen, sheep, and hogs; in all, 196 ani- 
mals. Jour times the number of oxen is equal to the num- 
ber of sheep, plus the number of hogs, minus 16. Four 
times the number of hogs is equal to the number of sheep, 
plus the number of oxen, plus 4. How many has he of each 
kind? Ans. He has 36 oxen, 120 sheep, and 40 hogs. 


19. Four boys bought a foot-ball for $2.00. The first 
paid $1.00 less than the other three together. Two-fifths 
of what the first and second paid, plus 35 cents, is 3 of what 
the third and fourth paid. Two-thirds of what the first 
three paid, minus 65 cents, is 5 of what the fourth paid. 
What did each pay? Ans. Each paid 50 cents. 


20. The population of four towns amounted to 7000. The 
population of the first and third together was equal to that 
of the second and fourth together. The population of the 
first and second together was 1000 less than that of the third 
and fourth together. The population of the first and fourth 
together was 2000 more than that of the second and third 
together. What was the population of each? Ans. The 
population of the first town was 2000; of the second, 1000; 
of the third, 1500; and of the fourth, 2500. 





172 ELEMENTARY ALGEBRA. 


SECTION XLVII. 


Peculiar Equations with more than Two 
Unknown Quantities. 


Prosiem 1. Given 4 ~ ty + w= 22 to find the 


values of x, y, z, and w. 


Sotution. It may be observed that one letter is wanting 
in each equation. There is no w in the first equation, no z 
in the second, no y in the third, and no 2 in the fourth. 


Adding the corresponding members of all the equations, 
8x2+ 38y+ 3824+ 38w = 90. 
Dividing by 3, “ ety+tz2+uw = 30. 


Now by subtracting the members of each of the original 
equations separately from the members of this last, the 
values of the unknown quantities can be found. Thus: 


xty+z+w= 30. 
x+y+z =. 21. 


9; 
cety+tz+w= 30. 
x+y + w= 22. 
Zz ra 
ety+tzt+w= 380. 
x +z+w = 23. 
y asi Ji 
xatytz+w= 30. 
ytez2t+u= 24. 


x = 6, 





Ww 





ELEMENTARY ALGEBRA. 173 


1 1 1 
Zhao anes ede ome VA Ud 
Big goeat tot Ne: 

; 3 RAD £9 | 

PRoBLEM 2. Given < —- + —-+ -= 34 7 to find the 

Bi tt eee 
4° § 2.3 
Gi gueae s 


values of x, y, and z. 


Soturion. Here it will be seen that clearing of fractions 
would give quantities containing zz, ry, yz, and xyz, making 
the solution difficult. The equations can be best solved 
without clearing of fractions till an equation is found with 
only one unknown quantity. Multiplying both members 
of the first equation by 4, we have 





A ae AS 
Soy ee ae 
The second equation is 2 -- -}- semees tt 
tpi RIERA AS 2 
: i gay 5 
Subtracting, - ie ; = = 
Sar. 5G. Oe 
The third equation is aD f na ae 
Multiplying both members of 3 4 3 rp 3 13 
the first by 3, py cy ah PAs 
Thriaid 13 
Subtracting, S83 15 
Now, taking the two equations that contain x and y¥ only, 
this iis 
age api $9: 
i Lae a 5 
. fon 
Subtracting, es 
Clearing of fractions, a 
Or, y= 4, 


15* 


174 ELEMENTARY ALGEBRA. | 


Substituting 4 for y in the equation : + : = a we have, 
Lie 5 
Pay bs: 
T ; 1 op 2 
ransposing, 5 oR ee 
ee? 
Bic 
Clearing of fractions, a= Mel 
Or, v= 38: 


Substituting the values of x and y in the first equation, 
lL) doa 


we have Siar + ete) 

Transposing 1 - 

: z° : 1295050 

1 ae 

2 aan 

Clearing of fractions, Das eee 

Or, z=s 2, 
xyz = 40 

Prosuiem 38. Given oh a a to find the values of 

yzw = 8 


X,Y, 2, and w. 


Soxtution. Multiplying the corresponding members of 
both equations together, we have, «x*y*z’w* = 64000. 
Extracting the cube root of 


both members, axyzw = AO. 
Dividing by the members of 

the first equation, es 
Dividing by the members of 

the second equation, Z == 2 
Dividing by the members of 

the third equation, y = 4, 


Dividing by the members of 
the fourth equation, 7 2s 


ELEMENTARY ALGEBRA. 175 


EXAMPLES. 


Find the values of the unknown quantities in the follow- 
ing sets of equations. 
(atytz+itutv+w= 28. 
etytz+tt+utv—wel4 
etyt+z+titu—v= 9. 
l<extytz2+t—u=d5. 


ety sao 

a+ty—z= 0. 

z2—y= —l. ) 

An. @ =1l;y=2;2=38;t=4;u=5;17=6; 
w= 7. 


Notsr. Subtract the members of the second from those of the first, 
and find w. Substitute that value in the second, and transpose. 
Then with the resulting equation and the third, the value of v can 
be found in the same way as that of w. Continue in this manner 
until the values of all the unknown quantities are found. 


2x—8y+4z—59 + 6w= 18. 
2x—8y+4z2+ 54 + 6w = 68. 
2. —22+ 38y—42z2+ 6w= 5. 
—2e+38y+42= 29. 
38y + 2a = 37. 
Ans ©= 83; y=7;2=6; 7=565:;w=—A4, 


Bal 3 
Bos 1 1 
ae 3 Ang.:G = 4: y= 8; 2 =12. 
ee ek J 
ytz oA 
a sac 2 
ae | 
/ a pea 
fee BG Ans 2 ao ee’ 12% e515: 
Ba) 2. 25 
P-2 | 
ose ee 
5.42 +2 = 15. Ang, 0 ws Aes: 27 oD eg ee) 
Ria is 2 =. LA; 


176 ELEMENTARY ALGEBRA. 


Bice U 2 
aye teed - == 25 
Py 
1-9 5 
6. ity a: St 2. 
eee Si 
Pista tpt ee oh 
Ans. © = 8; y=9;z= 10 
PO ce 
at y 102 alee : 
3 2 (ieee nA 1 
hase 9g Caer 
nO i 10 ta ae 
Bh iy eee ee 
Ieee 
dhe: 


Ans. « = 24; y = 10; 2 = 103 @ es) 
Notre. Substitute : * for = in the first three equations. That is, 


12 4 3 
put = for As for 4 and = for E Then collect the terms con- 


ae x. After this subtract 5 times the members of the second 
equation from the members of the third, and find z. 


sx+4y+ 2= 81. 
2y+32—3t = 5. 

Be Eine 2H eal ati oO: 
je y +24 "16; 
ge i fa» 


Ans. © = 38; y = 4; 2 = Geter ee 

Nore. From the last equation, « — ¢—2. Substitute this in the 

third equation, obtaining ¢ = 3z—11. Substitute 3z—11 for ¢in 

the second and fourth equations. Then the first, second, and third 

equations will contain only z, y, and z as unknown quantities. 
Eliminate z. 


ectytz =12. 
g J@tytu= 13. 
“)ant2z2+u= 20. 


yt2+u= 21. 
An. (= 1; y=2;2=9;u= 10. 


ELEMENTARY ALGEBRA. 177 


xyz = 60. 
ney = 30. 
10. pea 6. 
uyz = 20. 


ARS i 3 hf 0 ee = ee 


SECTION XLVIII. 
Problems relating to Work. 


Prosuiem 1. A can perform a piece of work in 12 days; 
B can do it in 8 days: and C can do it in 9 days. In what 
time can the three perform it? 

Sotution. Let x represent the number of days in which 
all three can do it. 

Then 2 represents the part which all three can do in one 
day. | 

Now, by the conditions of the question, 


A can do - of it in one day. 


B can do : of it in one day. 


C can do : of it in one day. 


9 
1 beg 1 
Hence 15 oe 5 =; gus 
Clearing of fractions, 62+ 9x+ 8x= 72. 
3 W2=—= 72. 
c= O3%5- 


Hence, all three together can do it in 3,3, days. 


ProsuiEM 2. A, B, and C can fell a given quantity of 
timber in 4 days; A and B can do it in 6 days; B and C, 
M 


178 ELEMENTARY ALGEBRA. 


in 7 days. How long would it take each separately, and 
how long would it take A and C together ? 


Sotution. Let x represent the number of days in which 
A can fell the timber. 
_ Let y represent the number of days in which B can fell 
the timber. 7 

Let z represent the number of days in which C can fell 
the timber. 


Then : represents the part of it which A can fell in one 


day. 
So ; represents the part of it which B can fell in one 
day. 
And * represents the part. of it which C can fell in one 
day. 
By the conditions of the question, 
Le Se vat 
py ee eae 
oi Weis 1 
mp re ty orf? 
1 21 eet 
ty aa ik 
Subtracting the members of the second equation from 
| 1 1 
those of the first, we get me 
Clearing of fractions, 12°=a.e 
Or, 2 ceee ee 
; ete 1 
Subtracting the members of the equation ae from 
th f the third, we get ie 
ose of the third, we g | ee 
Clearing of fractions, 84 = 5y. 


Whence, y = 164. 


ELEMENTARY ALGEBRA. 179 


Subtracting the members of the third equation from those 


1 3 
f the first, t me a) ge a 
of the first, we ge | . 58 
Clearing of fractions, 28 = 8 2. 
Whence, wae Sy. 


Hence, A can fell the timber in 94 days, B in 164 days, 
and C in 12 days. 


To find out how long it would take A and C together, 
observe that A can fell 2 of it in one day, and C in of it. 
Ay 4 
12 21 
of it. To do the whole of it, it would take them as many 


Hence, A and C together can fell se + of it; that is 


as a ated | 
days as a1 8 contained times in 57° 


Hence A and C can do it in 51 days. 


which is 5} times. 


ProsBiEM 3. A and B work at a boat for 10 days; when 
they call in C; and the three finish it in 10 more days. 
Had C worked from the beginning, it would have been com- 
pleted in 1513 days. In what time could C alone do it? 


Sotution. Although the question only calls for the 
time in which C could do it, the time in which A and B 
could do it must be represented by some letter, in order to 
form an equation. It is not necessary to consider A’s and 
B’s labor separately, as they do not work separately. 

Let ¢ represent the time in which A and B together could 
do it. 

Let x represent the time in which C could do it. 


Then : represents the part of it done by A and B to- 
gether in one day. 


And : represents the part of it done by C in one day. 


180 ELEMENTARY ALGEBRA. 


By the conditions of the question, (1 meaning once the 


( = 1) ea 
whole work, ) om ss a 
1575 of 1515 meee 
t x ‘) 
9 300 300 
The second equation is equal to 19: + igaue 1. 


Multiplying the members of the 
first equation by 43, oer et ee 





19% ° “192s 

, 150 4 
Subtracting, i95 ee 
Reducing, and clearing of fractions, = 374. 


Hence C alone could finish the boat in 374 days. 


EXAMPLES. 


1. A can do a piece of work in 10 days; B, in 15 days; 
C, in 20 days. How long will it take all together? Ans. 
4.8. days. 

2. Smith can do a certain piece of work in 8 days; Jones, 
in 6 days; Brown, in 9 days. How long will it take them, 
all working together? Ans. 244 days. 

3. A and B can mow a certain field in 8 days; B can do 
itin 20 days. How long would it take A? Ans. 131 days. 


4, A and B can mow a certain field in 8 days. If C join 
them they can mow it in 6 days. How long would it take 
C? Ans. 24 days. 


5. A shoemaker can fill a certain order in 3 weeks. His 
apprentice could fill it in 4 weeks. How long would it take 
both together? Ans. 13 weeks. 

6. Mr. Smith and his oldest son can do a piece of work in 
8 days. He and his youngest son can do it in 10 days. 
The two sons together can do it in 24 days. In what time 
can each do it alone? Ans. Mr. Smith can do it in 10}9 


ELEMENTARY ALGEBRA. 181 


days; his oldest son, in 30 days; and his youngest son, in 
120 days. 

7. A and B can do a piece of work in 93 days; A and C, 
in 113 days; and Band Cin 15 days. In what time can 
each do it alone? Ans. A can do it in 16 days; B, in 24 
days; and GC, in 40 days. 

8. A cistern is supplied by two pipes. One can fill it in 
1 hour, the other in 50 minutes. In what time will both 
running together fill it? Ans. 27,3 minutes. 


Notre. 1 hour — 60 minutes. 


9. A can do a piece of work in 7 days; B, in 8 days; C, 
in 14 days. In what time can all three together do it? 
Ans. 218 days. 

10. A cistern has three inlets. By the first it can be 
filled in 15 minutes; by the second, in 174 minutes, and by 
the third in 221 minutes. In what time will it be filled by 
all three together? Ans. 522 minutes. 


11. James can dig a certain sized trench in 14 days; his 
father can do it in 5 days; his older brother in 7 days. 
How long would it take all together? Ans. 242 days. 


12. A and B can perform a piece of work in 163 days. 
They work together for 10 days, when A leaves B to finish 
the work, which he does in 20 more days. In what time 
could each do it separately? Ans. A could do it in 25 
days; B, in 50 days. 

Nore. Let z— the number of days in which A would do it. . 

Let y = the number of days in which B would do it. 


Then : = the part done by A in one day. 
L 


And f = the part done by B in one day. ° 
eae 
: 1 1 1 
By the conditions of the question, - I ean 16% 
And, since 1 time the work stands for the 
‘ 10 , 30 
whole of it, i + 7) pone 


16 


182 ELEMENTARY ALGEBRA. 


13. A could reap a field in 10 days. If B assisted him 4 
days, it could be reaped in 5 days. How long would it take 
B to reap it? Ans. 8 days. 

14. A can do a piece of work in 8 days; B, in 6 days. 
If they both work at it for 3 days, C can finish it in 2 days. 
How long would it take C to do it? Ans. 16 days. 

15. A and B work at a wall for 7 days, and then call in 
C. The three finish the work in 7 more days. A and B 
could build it in 16 days. B and C could build it in 15 
days. In what time could each do it alone? Ans. A could 
do it in 7334, days; B, in 2029 days; and C, in 56 days. 

16. A and B can do a piece of work in 20 days. They 
do 2 of it, and C assists them with the remainder, the work 
being done in 17 days. In what time could C alone do it? 
Ans. 334 days. 

17. A can mow a field in 12 days. B and C together can 
mow it in 7 days. If all three work for 4 days, C can finish 
it in 3 days. In what time can B or Cdoit? Ans. B can 
do it in 9 days; ©, in 31} days. 

18. A and B can do a piece of work in 5 days. <A could 
do 4 of it in 3 days. How long would it take B to do the 
work? Ans. 114 days. 

Notr. First find how long it would take A to do the whole. 


19. A can do half of a piece of work in 9 days. B can 
do 4 of it in 7 days. How long would it take both together 
to do the whole. Ans. 9,9; days. p= 

20. A and B can do { of a piece of work in 2 days. B 
can do 4 of the work in§ days. How long would it take 
A to do half of the work? Ans. 74 days. 





ELEMENTARY ALGEBRA. 183 


SECTION XLIX. 


Ratio and Proportion. 


oe : : 246 
A ratio is a fraction. Thus, the ratio of 6 to 2 is =; and 


: i SEU. : 
the ratio of ato bis;. Since= = 3, We may say, the ratio 


b 2 
of 6 to 2 is 3. 
A ratio is generally represented thus, 6:2, which is the 
5 So a: 6 is the same as - 
The first term of a ratio is called the antecedent; the 
second term is called the consequent. Thus, in the ratio 


c+a 
n 


same as 6 -— 2, or 





, the antecedent is ¢ + a and the consequent is n. 


A proportion is an equality of ratios. Thus, : = 2 It 
may also be written, 3:4 = 6:8, or 3:4::6:8, All 
these expressions mean the same thing. 

A proportion must have four terms, since each of its 
ratios has two. The first and the last terms of a proportion 
are called its extremes; the other two are called its means. | 
In the preceding proportion, 3 and 8 are the extremes, and 
4 and 6 are the means. 

When the means of a proportion are the same, each of 
them is called a mean proportional between the other two, 
Thus, in the proportion, 4:8 :: 8:16, there is a mean pro- 
portional; it is 8. In the proportion, 4:3 :: 8:6, there is 
not a mean proportional. 


If we find that two fractions are not equal, they do not 


form a proportion. Let us try if 8:4::2:5 is a true 


3 2 
proportion. Set the ratios down in this way, A and 5 


These fractions are not equal, so the proportion is not a 
true one. Let us try 12:15::16:20. Set the fractions 


184 ELEMENTARY ALGEBRA. 


down thus, e and a These are equal, as may be found 


by reducing them to their lowest terms. Each will reduce 


4 5 ee 
to 5 Hence the proportion is true. 


QUESTIONS. 


Set down the proportion, 7:5:: 14: 10, in a different way. Is it 
a true one? 

Set down each of the following expressions in the form of an 
equation of fractions, and then try whether or not they are true 
proportions : 


a8 8 238 aes 
5:3 32 6210; 
b:O 2 1s 6 
Sa mated 1 ten BP 
ss tegey | kee ds Wika 
D SB al tu. 
pee Ra kee hs head hs 
m:n ::am:an 
G25 CY 3)2 02: 
G25 709 ets oY 
OF 298 Ts SEY 


sayst ay sys, 
Cb is “ea id 

In this last, do you know whether or not the proportion is true? 
Would it be true if the letters stood for the following numbers; a for 
8; bfor 5; cfor 6; d for 10? On what does the truth of a literal 
proportion sometimes depend? 

Select four of the above proportions, and name each antecedent 
and each consequent in them. Name also their extremes and their 
means. 

If any of the following proportions contains a mean proportional, 
point it out: 


ELEMENTARY ALGEBRA. 185 


Theorem. 


In any proportion, the product of the extremes is equal to 
the product of the means. 


To prove this, let any proportion be represented by 


EU Eee cts 
Set it down thus, 
ae 
be ds 
Clearing of fractions, we have 
ad = be. 


This is what was to be proved. 


Proportions may be tested by the above theorem. Thus, 
in the following expression, 8:5::4:6, the product of 
the extremes is 18, and the product of the means is 20. 
Hence the expression is incorrect, or, in other words, 3 and 
5 are not proportional to 4 and 6. Let the learner test all 
the preceding proportions in this way. 


PROBLEM l. ie the value of x which will make true 
—62 22 aa apd | 
4 ; 3 a a 


Soxtution. Since the product of the extremes is equal to 
ie OTe ae 2 3 





the proportion — 





the product of the means, 





| Pe 
Clearing of fractions, 2% —12% = 2x —3. 
Cancelling 22, —12« = —3. 
Or, 12% = 3. 
3 1 
Paley Bs 


ProspitemM 2. A and Bare merchants. A gains $5000, 
and B loses $3000. Then A’s capital is to B’s as 26 is to 
12. Had A lost $3000 and B gained $5000, A’s would have 


been to B’s as 17 to 20. What is the capital of each? 
16* 


186 ELEMENTARY ALGEBRA. 


Sontutrion. Let x represent A’s capital; and let y repre- 
sent B’s capital. 
Then, by the conditions of the question, 
18 + 5000 : y — 8000 :: 25: nf 
x — 3000: y + 5000 :: 17: 20 
Making the product of the extremes qu to the product 
of the means, 


122 + 60000 = 25 y — 75000. 
20 x — 60000 = 17 y + 85000. 


Bringing these equations into the proper form for elim- 
ination, 


‘ae —By= Tye 


202—-17y= 145000. 
Multiplying the mem- 
bers of the first by 5, 60 « — 125y = — 675000. 
Multiplying the mem- 
bers of the second by 3, 60%— 5ly= 485000. 
Subtracting the upper 
members from the lower, 74y = 1110000. 
ie= 15000. 


Substituting in the equation 202 —17 y = 145000, 
20 x — 255000 = 145000. 
20 x = 400000. 
x t= ~ 20000: 
Hence, A’s capital was $20,000, and B’s $15,000. 


PRoBLEM 38. Divide 96 into two parts in the ratio of 5 
to 7. 


: Rai y meal) 
SouuTion. Since the ratio mee is the same as the ratio 7 


(x 
we may represent the two parts by 5x and 72. 
Then we have the equation, 52+ Tx = 96. 
12x = 96. 
c= 8. 


Hence, the parts are 5 times 8, and 7 times 8; or 40 and 
56. 


ELEMENTARY ALGEBRA. 187 


Note. A longer solution would be to represent the parts by z and 


ie =a. 26. 
y, giving 13 MeN 7. 


Making the product of the extremes in the proportion equal to the 


product of the means, zy = 96. or 2. opie mh tes As 
(a= 5y. fz—b5y= 0. 


By solving these equations, we should find z = 40 and y = 56. 
Still another solution would be to represent the parts by z and 


96 — zx, giving the proportion, x: 96 —ax::6:7. 
Making the product of the extremes equal to the product of the 
means, we have 7x = 480 — 5x. 


By solving this equation, we should find z = 40, and 96 —z = 
56. 


EXAMPLES. 
1. Given w : 20 :: 6: 12, to find the value of z. 

Ans. x = 10. 

2. Given fs a eet to find the values of x 

and y. Ans. 2 = 4; y = 7. 
ety:l+2:: 5:2 

8. Given 4 r+ 8y:384 2:: ve 4 + to find the values 
ox +y:b+2: =O 

of x, y, and z. ie eee a a 


4, Divide 100 into three parts, in the ratios of 4, 5, and 
11. Ans. 20, 25, and 55. 
Note. Represent the parts by 42, 52, and 11 z. 


5. Divide 87 into three parts, in the ratios of 8, 9, and 12. 
Ans. 24, 27, and 36. | 


6. The sum of two numbers is 90, and the less is to the 
greater as 7 is to 11. What are the numbers? Ans. 35 
and 55. 


7. Divide 96 into four parts, in the ratios of 4, 5, 7, and 
8. Ans. 16, 20, 28, and 32. 


8. A road from Baltimore to Boston passes through Phil- 
adelphia and New York. ‘The whoie length of the road is 
413 miles. The distance from Baltimore to Philadelphia is 
to the distance from Philadelphia to New York, as 10 is to 


188 ELEMENTARY ALGEBRA. 


9. One-thirtieth of the distance from Philadelphia to New 
York added to 21 the distance from Baltimore to Philadel- 
phia equals the distance from New York to Boston. How 
far are the cities apart? -Ans. Baltimore and Philadelphia, 
100 miles; Philadelphia and New York, 90 miles; New 
York and Boston, 223 miles. 


9. Find two numbers in the ratio of 7 to 11, such that if 
12 be added to each they will be in the ratio of 17 to 25. 
Ans. 56 and 88. | 


10. Find two numbers in the ratio of 5 to 7, such that if 
10 be subtracted from each they will be in the ratio of 3 to 
5. Ans. 25 and 35. 


11. The ages of two brothers are 30 and 35 years. In 
how many years will they be in the ratio of 9to 10. Ans. 
15 years. 


12. The ages of two brothers are 28 and 30 years. How 
many years have passed since they were in the ratio of 2 to 
3? Ans. 24 years. 


13. Find two numbers in the ratio of 4 to 3, which being 
increased respectively by 10 and 30, will be in the ratio of 
js tod. Ans. 30 and 45. 


14. Find two numbers in the ratio of 3 to 4 whose sum is 
49. Ans. 21 and 28. 


15. Find two numbers such that if 12 be added to each, 
the results may be in the ratio of 6 to 7, while if 12 be sub- 
tracted from each, the results may be in the ratio of 2 to 3. 
Ans. 24 and 80. 


16. A loses $1000 and B gains $2000, when A’s money is 
to B’s as 2 is to 3. Then A gains $2000 and B loses $3000, 
when A’s money is to B’s as 8 is to 9. How much had each 
at first? Ans. A had $15,000; and B $19,000. 


17. If a certain lot of ground had been 12 feet longer, 
and 12 feet wider, the length would have been to the breadth 
as 7 to 6. But if it had been 12 feet shorter and 12 feet 


ELEMENTARY ALGEBRA. 189 


narrower, the length would have been to the breadth as 5 
to 4. How long and how wide was it? Ans. 72 feet long, 
and 60 feet wide. 


18. A farmer rents a farm for $247 per annum, $3 being 
reckoned for each acre of cultivated land, and $2 for each 
acre of pasture. The number of cultivated acres is to the 
number in the whole farm as 5 to 7. How many acres are 
there of each? Ans. 65 acres cultivated, and 26 acres of 
pasture, 

19. The price of a pound of a certain quality of sugar 
wants 2 cents to be to the price of a pound of the best 
raisins as 2 to 3. 9 pounds of the sugar equal in value 4 
pounds of the raisins plus 12 cents. What is the price of 
each? Ans. Sugar, 8 cents a pound; and raisins, 15 cents 
a pound, 

20. Four places, A, B, C, and D, are situated in the same 
order as the letters. From A to Dis 110 miles. The dis- 
tance from A to B is to the distance from B to C as 6 to 7. 
The distance from C to D added to twice that from A to B 
equals three times that from B to C. What are the dis- 
tances? Ans. From A to B, 30 miles; from B to C, 35 
miles; and from C to D, 45 miles. 


91. Find two numbers in the ratio of 9 to 5 whose sum 
shall be to their difference plus 22 as 7 isto 3. Ans. 99 
and 56. 


22. If I take $25 from each of two purses, the quantity 
remaining in the first is 2 of that in the second. If I then 
take $20 more from each, the quantity remaining in the first 
is 7° of that in the second. How much was in each at first? 
Ans. $75 in the first, and $100 in the second. 


23. The first of four numbers is to the second as 2 is to 7, 
and the third is to the fourth as 14 is to 27. The sum of 
the first and third is to the sum of the second and fourth as 


190 ELEMENTARY ALGEBRA, 


2isto 5. The sum of all the four numbers is 77. What 
are they? Ans. 8, 28, 14, and 27. 
Norse. Represent the numbers by 22 and 72, 14y and 27y. 


24. The weights on one plate of a pair of scales are to the 
weights on the other plate as 5 is to 6. A part being taken 
off each plate, the weights remaining exactly balanced each 
other. The whole amount of weights remaining was 30 
ounces. The parts taken off were in the ratio of 5 to 7. 
What were the weights at first? Ans. 30 ounces and 36 
ounces. 


_ Notr. Represent the weights at first by 5z and 6z. Take 15 
ounces from each to get the parts taken off. 


25. Find three numbers whose sum is 38, such that if 2 
be subtraeted from the first and from the second, the remain- 
ders will be in the ratio of 3 to 5; and if 4 be subtracted 
from the first and from the third, the remainders will be in 
the ratio of 2 to 7. Ans. 8, 12, and 18. 

26. Find three numbers whose sum is 47, 4 of the first be- 
ing to 1 of the second as 4 is to 3, and 4 of the second to , 
of the third as 5 is to 2. Ans. 12, 15, and 20. 


27. From two boxes of sugar were taken quantities in the 
ratio of 5 to 7. If 6 pounds less had been taken from the 
former, only } as much would have been taken from it as 
from the latter. How much was taken from each? Ans. 
20 pounds from the first, and 28 pounds from the second. 


28. There are two numbers in the ratio of 3 to 4. If 10 
be added to each, the sums will be in the ratio of 5 to 6. 
What are the numbers? Ans. 15 and 20. 

29. Divide the number 36 into two such parts that the 
quotient of the greater part divided by 6 shall be to the quo- 
tient of the less divided by 30 as 25 isto1. Ans. 30 and 6. 

30. Find two numbers in the ratio of 5 to 6, such that if 
each be increased by 12 they will be in the ratio of 11 to 
12. Ans. 10 and 12. , 


ELEMENTARY ALGEBRA. 191 


31. In a school for boys and girls, if there were 12 more 
of each, there would be 6 boys to every 7 girls; and if there 
were 12 less of each, there would be 4 boys to every 56 girls. 
How many are there of each? Ans. 60 boys and 72 girls. 


32. A certain piece of brass weighs 18 pounds, and con- 
tains 2 pounds of copper for every pound of zinc. How 
much copper must be added to make a kind of brass contain- 
ing 5 pounds of copper for every 2 of zinc? Ans. 3 pounds. 

Nore. There are now 12 pounds of copper and 6 pounds of zine. 
Let x represent the amount of copper to be added. Then, z+ 12 : 
Be: bd 5 2. 

33. John’s marbles were to James’s as 6 isto 7. John 
won 8 from James, and then their marbles were as 8 to 5. 
How many had each at first? Ans. John had 24, and 
James had 28. 


34, Given 9: x::3: 4, to find-the value of z. 


Ans. x = 12. 

35. Given «:6::8: 4, to find the value of x. 
Ans. x = 12. 

36. Given 5:3:: x: 12, to find the value of z. 
Ans, x = 20. 

37. Given 2: 11::6: 2, to find the value of z. 
Ans. x = 33. 
Bogen} to! on 4 4.5 f00,, find the 
values of x andy. _ Ans. & = 44 = 2 
39. Given 15 Lhe me , 3} to find the values of «x 
and y. Ans, 2 =. 5; y = 2, 


eatyt4: w«—y+2::138:9 
40, Given Tty—2z: w2—y—z:: 5:1} to 
—(—aty—z:—ax—yst2z:: 9:5 
find the values of x, y, and z. 
Ansiie se] vp = 2 eee} 


192 ELEMENTARY ALGEBRA. 


SECTION L. 
Arithmetical Progression. 


In the series of numbers, 1, 3, 5, 7, 9, 11, every term 
after the first is formed by adding the same number to the 
term before it. This series is called an arithmetical progres- 
sion; its common difference is 2. 

In the series, 17, 14, 11, 8, 5, the common difference is 
negative. It is —3 that we add to 17 to make 14; and so 
on. 

An arithmetical progression is a series of terms which in- 
crease or decrease by a common difference. 

The following are examples. Let the learner tell the common 


difference in each series. It may always be found by subtracting 
a term from that which comes after it. 

4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74. 

81, 72, 68, 54, 45, 86, 27, 18, 9, 0. 

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, &c. to any extent. 

3, ay td, 15. 

0, 10, 20, 30, 40, 50. 

16, 14, 12, 10, 8, 6, 4, 2, 0, —2, —4, —6, —8. 

— 5, — 3, —1, 1, 3, 5, 7, 9. 

—5, —7, —9, —11, — 18, — 15, —17. 


Finding Terms of an Arithmetical Progression. 


Any arithmetical progression may be represented by let- 
ters. We may represent the first term by a, and the com- 
mon difference by d. Then any arithmetical progression 


will be aoa+d,a+ 2d,a+ 3d,a + 4d, &e. 


Thus we see that to obtain the second term, we add once 
the common difference to the first term; to obtain the third 
term, we add twice the common difference to the first term ; 
to obtain the fourth term, we add three times the common 
difference to the first term; and, in general, to obtain any 
term, we take the common difference one time less than the 


ELEMENTARY ALGEBRA. 193 


number of the term, and add it to the first term. In a de- 
creasing series, it is a negative quantity that we add. 

Let / represent the last term of any arithmetical progres- 
sion, and n the number of terms. Then, from what we have 


said, Fata He lh 


ProsLEM 1. The first term of an arithmetical progres- 
sion is 4, and the common difference is 8. What is the 9th 
term ? 


SotutTion. Here the 9th term is the last term of the 
series as far as we consider it; so that a = 4,n = 9, andd 
= 8, in the formula ~=a+(m—1)d. 

Hence, ~= 4+ (9—1) x 8. 
Performing the operations, / = 68. 


ProspLEeM 2. In a decreasing arithmetical progression, 
the first term is 100, the common difference is — 5, and the 
number of terms is 20. What is the last term ? 


SoLUTION. ~=a-+ (n—1)d. 
= 100 + (20 —1) x —5. 
Performing the oper- 
ations, i = 5. 


EXAMPLES. 


1. In an arithmetical progression, the first term is 9, the 
common difference is 9, and the number of terms is 9. Re- 
quired the last term. Ans. 81. 

2. In an arithmetical progression, the first term is 27, the 
number of terms is 6, and the common difference is — 2. 
What is the last term? Ans. 17. 

3. A merchant paid 12 creditors sums of money in arith- 


metical progression. He paid the first $100; the second, 
$150; and soon. What did he pay the last? Ans. $650. 

4, A body falling without any hindrance will fall 16,4 
feet in the first second, 484 feet the next, and so on in arith- 


metical progression, the common difference being 32} feet. 
LT N 


194 ELEMENTARY ALGEBRA. 


How far will it fall during the seventh second? Ans. 2097, 
feet. 

5. Given a = 7, andd = 5. Required the 8th term. 
Ans. 42. 

6. Given a = 4, and d = 100. Required the 5th term. 
Ans. 404. 


7. Given a = 0, and d = 24. Required the 3d term. 
Ans. 48. 


8. Given a = 100, and d = 21. Required the 21st term. 
Ans. 150. 

9. Given a = 2}, and d = 74. Required the 17th term. 
Ans. 1223. 

10. Given a = 3}, and d = 4. Required the 20th term. 
Ans. 794. 


11. Given a = 8, andd = —2. Required the 4th term, 
Ans. 2. 

12. Given a = 29, and d = —3. Required the 8th 
term. Ans. 8. 

13. Given a = 72, and d = —4. Required the 19th 
term. Ans. 0. 

14. Given a = 60, and d= —10. Required the 12th 
term. Ans. — 50. 

15. Given a = 15, and d = —7. Required the 6th 
term. Ans. — 20. 

16. Given a = —10, and d = —3. Required the 21st 


term. Ans. — 70. 


17. Given a = 4, andd = 2. Required the 5th term. 
Ans. 12. 


18. Given a = —8, andd = 6. Required the 8th term. 
Ans. 39. 
19. Given a = —7, and d = —3. Required the 7th 


term. Ans. — 25. 


20. Given a = 5, andd = 8. Required the 4th term. 
Ans. 29. 


ELEMENTARY ALGEBRA. 195 


SECTION LI. 


Arithmetical Progression. (Continued.) 


Finding the Sum of the Terms. 


TAKE any arithmetical progression, and write under it 
the same series in an inverted order, adding together the 
corresponding terms. For example: 

Soo ATI a 8 eee I oad Eo Bene 8 
TANS dt Soe a ia eat bal ey AN i 
Lon eer. 205.620.5026... 26. 

It will be observed that the sums are all the same. This 
must always be so; for the terms of one progression increase 
just as much as the terms of the other decrease. 

Each sum is equal to the first term plus the last. Each 
sum may therefore be represented by a + J. 

There are as many sums as the original series had terms ; 
that is, there are n sums. Hence the sums altogether 
amount to (a + /) n. 

This amount is equal to the sum of the series taken twice. 
Hence, the sum of the series, which we may call s, is half 
that amount. That is, 


Propiem 1. The first term of an arithmetical progres- 
sion is 6; the last is 20; and the number of terms is 8. 
What is the sum of the series ? 





Soturion. In the formula, $= ae 
Substituting, c= a ) a) 


Performing the operations, s = 104. 


Note. If we want to find the common difference of this series, we 
must take the formula of the preceding section, 7 = a + (n —1) d. 


196 ELEMENTARY ALGEBRA. 


Hence, if we take a from J, we shall have (nm —1) d. 

As n = 8 in this example, taking a from J will give 7d. 

And, as / = 20 and a = 6 in this example, / —a is 20 —6, or 14. 
Hence, 7d = 14, whence d = 2. 


We may now write out the series, 
6, 8, 10, 12, 14, 16, 18, 20. 
By adding the terms together, we may prove the correctness of 


the preceding answer. The sum will be found to be 104, as it was 
found to be by the formula. 


PRoBuLEM 2. What is the sum of the first 500 odd num- 
bers. 


SoLuTIon. Here, as we wish to find the sum of an arith- 
metical series, we should know the first term, the last term, 
and the number of terms. Now, a = 1, and n = 500, but 
7s not given. Hence it must be found as in Section L. 

~=1-+ (500 —1) x 2. 








PERE 
Then, in the formula of the present section, 
leat gen 
c= cae 
(1 + 999) x 500 
Ce ee ——; 
2 
s = 250000. 
EXAMPLES. 
1. Given a = 20,1 = 76,n = 15. Required s. 
Ans. 720. 
2. Given a = 7,/ = 103,n = 16. Required s. 
Ans. 142. 


3. Givena = 0,7 = 5,n=11. Required s. 
Ans. 274. 


4, Given a = —10,/ = —20,n = 6. Required s. 
Ans. — 90. 
5. Given a = —4,/= 8,n = 5. Required s. 


Ans. 10. 


ELEMENTARY ALGEBRA, 197 


6. Given a = —9, 7 = —49,n = 10. Required s. 
Ans. — 290. 
7. What is the sum of the first 1000 even numbers? 
Ans. 1,001,000. 


8. What is the sum of the first 750 odd numbers? Ans. 
562,500. 

9. What is the sum of the first 2000 numbers, 1, 2, 3, 4, 
5, 6, &e.? Ans. 2,001,000. 


10. What is the sum of all the numbers from 101 to 300, 
inclusive? Ans. 40,100. 


11. How many strokes does a common clock strike from 
1 p.m. to midnight, inclusive? Ans. 78. 


12. What is the sum of the eleven numbers, 4, 7, 10, 13, 
and so on, ending with 34? Ans. 209. 


13. An unresisted falling body descends 16,}, feet in the 
first second, and increases 323 feet each second over the 
preceding. Through what space will it fall in 7 seconds ? 
Ans. 788,}; feet. 


14. If a man walks on 20 successive days, going 10 
miles the first day, and on each of the following days going 
half a mile more than on the day before, how far will he 
walk on the last day, and how far in all? Ans. 193 miles 
on the last day; 295 miles in all. 


15. A man bought a farm of 150 acres, paying for the 
first acre $5; for the next, $7; and so on in arithmetical 
progression. How much did the farm cost, and how much 
did it average per acre? Ans. $23,100 in all; $154 per 
acre. 

16. The first term of an arithmetical progression is 3, the 
common difference is 2, and the last term is 79. What is 
the number of terms; and the sum of the series? Ans. There 
are 39 terms; s = 1599. 

Nore. To find the number of terms, take the formula 1 = a 
+ (n—1) d; substitute the values of /, a, and d, and find the value 


of n, the unknown quantity. 
17 * 


198 ELEMENTARY ALGEBRA. 


17. The first term of an arithmetical progression is 3283, 
the common difference 1s — 7, and the last term is 2413. 
What is the number of terms, and the sum of the series? 
Ans. n = 100; s = 28481. 


18. The first term of an arithmetical progression is 900, 
the common difference is — 100, and the last term is —900. 
What is the number of terms, and the sum of the series ? 
Ans. There are 19 terms; s = 0. 


19. The first term of an arithmetical progression is 25, 
the common difference is 7, and the number of terms is 101. 
What is the sum of the oe Ans. 37875. 


20. Insert 6 arithmetical means between 9 and 37. Ans. 
1,15 21,420,020, and on. 


Notre. Here the common difference is to be found. The number 
of terms is 8 in all, the six means, and the two extremes. We have 
a= 9,1= 37, andn= 8. Substitute these values in the equation, 
1 =a - (n—1) d; and find the value of d, the unknown quantity. 


87 = 9 + (8—1) d. 


387 —9 = 7d. 
28) = Tid. 
4= d. 


When the common difference is found, it is easy to continue add- 
ing it so as to find the required terms. 


21. Insert 4 arithmetical means between 2 and 7. Ans. 
3, 4, 5, and 6. 

22. Insert 8 arithmetical means between 4 and 31. Ans. 
(preuleWea SR Gime ER eee hats bas: 

23. Insert 3 arithmetical means between 0 and 20. Ans. 
5, 10, and 15. 

24. Insert an arithmetical mean between 200 and 250. 
Ans. 225, 

25. Insert 9 arithmetical means between 4 and 34. Ans. 
4,440,718,5°165 19 992-95 25 Pander. 

26. In a straight line from a basket are 100 stone the 
first being 1 yard from the basket, and each succeeding stone 


ELEMENTARY ALGEBRA. 199 


1 yard from the one preceding it. How far would any one 
walk who should start from the basket, go to each stone 
separately, pick it up, and return with it to the basket be- 
fore going to another. Ans. 10,100 yards. 


27. A man gives his son $1 one week, and for the next 
ten weeks increases his gift of the week before by $13. 
How much does he give him inall? Ans. $99. 


28. During 12 hours wine leaks from a cask, the leak in- 
creasing in arithmetical progression, 2? of a pint for each 


succeeding hour. The first hour 4 of a pint leaked out. 
How much leaked out during the last hour? How much 


in all? Ans. 42 pints the last hour; 284 pints in all. 

29. The first term of an arithmetical progression is 95, 
the number of terms is 49, and the last term 35. What is 
the common difference? Ans. —1}. 

30. The first term of an arithmetical progression is 10, 
the common difference 3, and the last term 100. What is 
the number of terms? Ans. 31. 


SECTION LII. 
Geometrical Progression. 


In the series of numbers, 3, 6, 12, 24, 48, 96, every term 
after the first is formed by multiplying the term before it by 
the same number. This series is called a geometrical pro. 
gression: its ratio is 2. In the series, 8, 4, 2, 1, 3, 7, the 
ratio is 1. In the series 5, —25, 125, — 625, 3125, the 
ratio is —9. 

A geometrical progression is a series of terms in which 
each one after the first is formed by multiplying the pre- 
ceding one by a certain number, which is called the ratio. 


The following are examples. Let the learner tell the ratio in each 


200 ELEMENTARY ALGEBRA, 


series. It may always be found by dividing a term by that which 
precedes it. 


5, 10, 20, 40, 80, a 320, 640, 1280, 2560, 5120. 


9, 3,1, 3, 3 ay gy xte- 

4, 5, 64, 743. 

2, —2, 2, —2, 2, —2. 

81, —27, 9, —3, 1, —4, 4, —d, 
100, 200, 400, 800, 1600. 

10, 100, 1000, 10000, 100000. 


6, — 3, 3, —3, 3, — s.., 


? 3) i = 16 


Finding Terms of a Geometrical Progression. 


Any geometrical progression may be represented by let- 
ters. We may represent the first term by a, and the ratio 
by r. Then any geometrical progression will be 


a, ar, ar’, ar’, art, ar®, &e. 


Thus we see that to obtain the second term, we multiply 
the first power of the ratio by a; to obtain the third term, 
we multiply the second power of the ratio by a; to obtain 
the fourth term, we multiply the third power of the ratio by 
a; to obtain the fifth term, we multiply the fourth power of 
the ratio by a; and, in general, to obtain any term, we raise 
the ratio to a power less by one than the number of the re- 
quired term, and multiply it by a 

Let / represent the last term of a geometrical progression, 
and 7 the number of terms. Then, from what we have said, 


l= ar 


ProsiEeM. Given, the first term of a geometrical pro- 
gression, 4, and the ratio, 3, to find the ninth term. 


SotuTion. In the formula, 7 = ar*—1. 
[=a ee 
1 = 26244. 


Norse. Observe that 8 must be raised to the 8th power before 
multiplying by 4. 


ELEMENTARY ALGEBRA. 201 


EXAMPLES. 


1. What is the seventh term of a geometrical progression, 
whose first. term is 100, and ratio 2? Ans. 6400. 


2. Find the eighth term of the geometrical progression, 
4,12, 36, &. Ans. 8748. 


3. Given a = 7 and r = 10; find the fifth term. Ans. 
70000. 


4, Given a = 6 and r = 13; find the 6th term. Ans. 


729 
LG” 

5. Given a = 7 and r = —1; find the 18th term. 
Ans. —7 

6. Given a = 32 and r = —4; find the 12th term. 
Ans. — 4. 


7. A man once agreed to give some children 1 cent for 
each one that was but one year old, 2 cents for each one 
two years old, 4 cents for each one three years old, and so 
on, doubling for each year. To how much was one entitled 
who was fourteen years old? Ans. $81.92. 

8. A man had seven daughters. He promised to give the 
youngest $10 for a Christmas present, and each of the others 
3 times as much as her next younger sister. What was the 
oldest to receive? Ans. $7290. 


9. A man offered to sell 10 geese, provided the purchaser 
would take them all at the rate of 1 cent for the first, 4 
cents for the second, and so on in geometrical progression. 
The other, thinking the bargain an excellent one, agreed. 
What was the price of the last goose? Ans. $2621.44. 


10. Find the 7th term of a geometrical progression whose 
first term is 6, and whose ratio is 2. Ans. ;3845. 





202 ELEMENTARY ALGEBRA. 


SECTION LIII. 
Geometrical Progression. (Continued.) 


Finding the Sum of the Terms. 


TAKE any geometrical progression, as, 3, 6, 12, 24, 48. 
The ratio here is 2. Let s stand for its sum, so that (setting 
it down in inverse order) 


s= 48 + 244124643. 
Now multiply each term of this: series by the ratio, and 
subtract the sum just set down from the result. 
rs = 96 + 48 + 244 124 6 
= 48 + 24+ 1246-4 3. 


Ts —s = 96 — 3. 





Now 96 was obtained by multiplying the last term of the 
original progression by the ratio. Hence 96 is the same as 
lr. The first term, or 3, is represented by a. 





Hence, rs —s = lr —a. 
Factoring, (r—1)s =r —a. 
Late lr —a 
Dividing by r —1, 6 ee 


This same formula may be put in another form by multi- 
plying both terms of the fraction by —1, which does not 
alter its value. The result is 





pe ah Ha 8 
ee op ee 
Rearranging the terms in the numerator and in the de- 
: a—ltlr 
nominator $= . 
Lo 7 


These two formulas are precisely the same in value; for 
multiplying both terms of a fraction by the same number, 
—1, does not alter the value of the fraction. Hither form- 
ula may be used in any case; but it is most convenient to use 
the latter form when the ratio is less than 1. 


ELEMENTARY ALGEBRA. 203 


We have now had the following formulas for geometrical 





progression : 
First, l= ar -—, 
lr — a 
Second, ENE ESS oi) Megat he dr 
r—l 1—r 


ProsieM 1. Find the sum of the terms of a geometrical 
progression, whose first term is 5, its ratio 4, and the num- 
ber of terms 7. 


SotutTion. The last term not being given, it must first 
be found. 





Take the first formula, ENT To he 
t=5 x 4, 
== 20480. 
Then take the second formula, s = b = 
pra 20480 x 4—5 
ieieess en 
s = 273805. 


ProsiEM 2. Insert three geometric means between 3 
and 243. 


SotuTion. In the formula, av”—! = J, we know a and 
Zto be 38 and 248. We also know n: it is 5, since there 
are three means to be found, and two extremes given. 


Substituting, we have, 31rt = 248. 
Dividing by 3, f= Sl. 
Taking the square root, r= 9. 
Taking the square root again, r= 3. 


Knowing the ratio, the means are easily found to be 9, 
27, and 81. 


ProsieM 3. Insert one geometric mean between 8 and 
32. 


204 ELEMENTARY ALGEBRA. 


Soturion. Take the formula, are eth 
Substituting, 87" = 382. 

7 =, 

gn 


The mean is consequently 8 x 2 or 16. 


This geometric mean is also a mean proportional between 
8 and 32; for, when we say the three quantities 8, 16, and 
32 are in geometrical progression, we mean that the ratio 
which 8 bears to 16 is the same as the ratio which 16 bears 
to 32. Hence, we may make the proportion, 


SLO eel Gare, 


EXAMPLES. 


1. Find a mean proportional between 20 and 5. Ans. 10. 


bo 


. Find a mean proportional between 2 and 162. Ans. 


. Find a mean proportional between 5 and 125. Ans. 25. 
. Find a mean proportional between 9 and 16. Ans. 12. 


Oo me 


. Insert one geometric mean between land 16. Ans. 4. 


6. Insert two geometric means between 1 and 8. Ans. 2 
and 4. 


7. Insert three geometric means between 4 and 324. Ans. 
12, 36, and 108. 


8. Find the sum of 6 terms of the geometrical progression, 
4,20, 100, &. Ans. 15624. 

9. Find the sum of 3, 10, 19°, &c., to eight terms. Ans. 

10. A man had a horse offered to him at 1 cent for the 
first nail in its shoes, 2 cents for the second, and so on, 
doubling for every additional nail, there being 32 nails in 
all. He thought it was a good bargain; but, on computing 
the amount, found he had not money enough to pay for the 
horse. How much would it have cost at the rate offered? 
Ans. $42949672.95. 


ELEMENTARY ALGEBRA. 205 


11 to 20. Find the sums of the progressions in the ten 
examples of the preceding article. In the 7th, find the sum 
on the supposition that there was one child 1 year old, one 
2 years old, and so on, one child for each number of years 
up to 14 years inclusive. 


Answers. 
11. 12700. 16. 1385, 
a2 13120. 17. $163.83. 
Dt Gt « 18. $10930. 
14, 1995, 19. $3495.25. 
$5. 0, 20. 19229". 
eee ostel: 


SECTION LIV. 


Miscellaneous Questions producing Simple 
Equations. 


1. THE sums of 3 numbers taken two and two are 62, 46, 
and 58. What are the numbers? Ans. 37, 25, and 21. 


2. The sum of 3 numbers is 360. Half the difference of 
the first and second is 15. Half the sum of the second and 
third is 105. What are the numbers? Ans. 150, 120, and 
90. 


3. Find 3 numbers such that the first plus 4 of the second 
equals 90; the second plus 3 of the third equals 70; and 
the third plus 4 of the first equals 60. Ans. 70, 60, and 50. 

4, Divide 80 into two such parts that the greater divided 
by their difference may give 3 as a quotient. Ans. 48 and 
32. 

5. Divide 48 into 4 such parts that the first increased by 
3, the second diminished by 3, the third divided by 3, and 
the fourth multiplied by 3, shall give equal results. Ans. 
6, 12, 27, and 3. 

Notr. See Section XLI., Example 29, 

18 


206 ELEMENTARY ALGEBRA. 


6. Four equal sums of money were lying on a table. $20 
were taken from the first; $30 from the second; $40 from 
the third; and $50 from the fourth ; when $260 were left 
on the table. What was the value of each sum? Ans. 
$100. 

7. A farmer owns 400 acres of land. He allows 1 acre 
of pasture for every 9 sheep he owns, and cultivates the 
rest of his ground. The number of acres under cultivation 
is 4 of the number of sheep he owns. What is that num- 
ber? Ans. 900. 


8. Divide 102-into two such parts that 56 subtracted from 
the greater shall leave the same remainder as when twice 
the less is subtracted from 76. Ans. 72 and 30. 


9. Divide 81 into three parts such that | of the first, 2 of 
the second, and 3 of the third, shall be equal. Ans. 18, 27, 
and 36. 


10. A merchant bought four books of entry. The ledger, 
journal, and blotter cost him $9.50. The journal, blotter, 
and cash-book cost him $7. The blotter, cash-book, and 
ledger cost him $8. The cash-book, ledger, and journal 
cost him $8.50. What did each cost? Ans. The ledger, 
$4; the journal, $3; the blotter, $2.50; and the cash-book, 
$1.50. 

11. A boy having 60 cents could buy 9 tops and 6 mar- 
bles, or 10 marbles and 5 balls, or 4 balls and 5 kites, or 6 
tops and 6 kites. What was the price of each article? 
Ans. Tops, 6 cents.each; marbles, 1 cent each; balls, 10 
cents each; and kites, 4 cents each. 


12. A person was desirous of giving 10 cents to each of 
some beggars, but had not enough change by 20 cents. He 
gave each of them 7 cents, and had 7 cents remaining. 
How many beggars were there? Ans. 9. 

13. Fritz and Hans can drink a cask of beer in 5 days. 
Fritz could drink it in 7 days. How long would it last 
Hans? Ans. 174 days. 


ELEMENTARY ALGEBRA. 207 


14, A steamboat that can go 18 miles an hour with the 
current, and 12 miles an hour against it, requires 14 hours 
to go a certain distance and return. What is that distance? 
Ans. 9 miles. 


Notre. Let x = the distance. It goes with the current, a mile in 


a5: 2% 
73 of an hour. To goz miles, takes 8 of an hour. So, to go z miles 


#2 
against the current, takes Dp ofan hour. The sum of these 2 times 
is 14 hours. 


15. What number equals 3 of ;*; of ;, of itself, plus 1253. 
Ans. 1280. : 


16. A man walks at the rate of 2 miles an hour for a cer- 
tain distance, and immediately returns on horseback at the 
rate of 12 miles an hour. His journey occupies 103 hours. 
How far does he go? Ans. 18 miles. 


17. One hundred pounds of boiling water will dissolve 40 
pounds of salt. How many pounds of boiling water must 
be added, so that the salt contained in 140 pounds of the 
mixture may be but 8 pounds. Ans. 560 pounds. 

Note. The mixture of salt and water is at first 140 pounds. Let 
x = the quantity to be added. Then the mixture becomes 140 + 
pounds, in which there are 40 pounds of salt. One-fifth of that much 
salt, or 8 pounds, will be contained in ¢ of the mixture. Now i of 


© 
140 + wis 28 + 5 This is what the question gives as 140 pounds. 


18. A does 1 of a piece of work in 6 days. B then as- 
sists him, and the two finish it in 8 more days. In what 
time could each do it alone? Ans. A could do it in 18 
days; B, in 36 days. 

19. Find a fraction which will equal 4 if 2 be added to 
its numerator, and 1 if 3 be added to its denominator. 
J a 

20. A merchant mixed some sugar that cost 6 cents a 
pound with some that cost 8 cents a pound. The mixture 


208 ELEMENTARY ALGEBRA. 


contained 200 pounds and had cost him $15. How many 
pounds did he take of each kind? Ans. He took 50 pounds 
at 6 cents, and 150 at 8 cents. 


21. What is the number whose third part Sheed its 
fourth part by 4? Ans. 48. 


22. A father’s age is 5 times that of his son. In 6 years 
his age will be only 3 times that of his son. What is the 
age of each? Ans. The father’s age is 30 years; the son’s, 
6 years. 


23. A man gave $1 to two poor persons, giving one 50 
cents more than'the other. How much did he give to each? 
Ans. He gave 75 cents to one, and 25 cents to the other. 


24. A tailor agreed to make a certain number of coats 
for $10 each. He made 5 more than he agreed to, and re- 
ceived for all $170. How many did he agree to make? 
Ans. 12. 


25. Two persons’ ages are to each other as 8 to 5. Fif- 
teen years ago they were in the ratio of 5 to 2. What are 
they? Ans. 40 years and 25 years. 


26. One pipe will fill a certain cistern in 4 hours, and 
another in 6 hours. How long will it take them to fill it, 
if both run together? Ans. 2 hours 24 minutes. 


27. A liquor-dealer has wine worth $5 a gallon, and wine 
worth $8 a gallon, and wishes to make a mixture of 120 
gallons worth $6 a gallon. How much of each must he 
take? Ans. He must take 80 gallons at $5, and 40 gallons 
at $8. 


28. Find two numbers such that the first increased by 9 
shall be 3 times the second, and the second increased by 3 
_ shall be 4 of the first. Ans. 36 and 15. 


29. If 32 pounds of sea water contain 1 pound of salt, 
how much fresh water must be added to these 32 pounds in 
order to make the quantity of salt contained in 32 pounds 
of the new mixture only 1 ounce. Ans. 480 pounds. 


ELEMENTARY ALGEBRA. 209 


30. There is a certain number of two digits. The sum 
of its digits is 11. If 19 be added to the first digit, the 
sum will be 5 times the second. What is the number? 


Ans. 65. 


SECTION LV. 


Involution. 


Involution is the formation of powers. 


The second power of a is a’, which means aa, or a multi- 
plied by a. The third power of a is a’, which means aaa. 


When we write a, it is the same thing as if we wrote a’. 

To raise a quantity whose index is 1 to any power, we 
have only to write the index of that power in the proper 
place. 


Let us now raise a’ to the fifth power. That is, let us 
multiply together a’, a’, a’, a’, and a*. By the principles 
of multiplication, we know that the product is a. That is, 
adding all the indices is the same thing as multiplying the 
index 3 by 5. In the same way, a‘ raised to the seventh 
power is a®, 


Let us next raise 3 a’x’y’z to the fifth power. Every fac- 
tor of this expression is used as a multiplier when we mul- 
tiply by the whole expression. Hence, we have but to find 
the fifth power of each of the factors, and write them to- 
gether for the fifth power of the whole. We find the fifth 
power of 3 to be 243, and write after it the fifth power of 
each of the others. The result is 243 a®x*y2°, 

When a negative quantity is to be raised to any power, 
the rule for the signs in multiplication must be observed. 

Let us raise —a to the seventh power. We know that 

18 * O 


210 ELEMENTARY ALGEBRA, 


—ax—a4=+¢@ 
—ax+ae=—da 
—ax—a= + a. 
—ax+t+a=— a’. 
—ax—a=-4 a’ 
—ax+t+a’= —d. 


Here we see that when we involve a negative quantity, 
even powers will be +, and odd powers will be —. 
To involve a fraction, involve both of its terms. 


PROBLEM. Raise a + 6 to the fourth power. 


SotuTion. This means, multiply a + b by a + 6; mul- 
tiply that product by a + 6; and then multiply that pro- 
duct by a + b. The operation is as follows: 


a + b 

a + b 

at ab 
ab+ @C 

a’ + 2ab-+ 6’ = the second power. 
a + 5b 





e+22¢b+a b? 
vb + 2a? + 5 
a@+3ab+3ab?+ 06° = the third power. 
Goer eh 
a+ 30°90 + 3070? + = ad? 
ab + 807b? + 3ab* + Ot 
at + 4a%> + 6 a°b? + 4ab* + b¢ = the fourth power. 





So to raise any polynomial to any power, take that poly- 
nomial as a factor as many times as the exponent of the 


power indicates. 
EXAMPLES. 


1. Square 3 a‘e; that is, raise it to the 2d power. 
Ans. 9 a®c?. 


2. Square — 3 ate. | Ans. 9 a®c*, 


21. 
Ans. x? + 2xy +7? = the 2d power. 


ELEMENTARY ALGEBRA, 211 


. Cube 7 a’xyz*; that is, raise it to the 3d power. 
Ans. 3438 abc? y2"4, 
. Cube —7 a’ayz’. Ans. — 348 abr? y*z”4, 
. Raise 8 ac’x*y*z* to the 4th power. 
Ans. 4096 ate®x?y'62™, 
. Raise —2acx to the 5th power. Ans. — 32 a’cbx', 
. Raise 2 acx to the 5th power. Ans. 32 aécia°, 
. Raise 7 m to the 2d power. ° Ans. 49 m’. 
. Raise a to the 27th power. Ans. a™. 
. Raise — a to the 27th power. Ans. —a”™, 
. Raise — 2 xty*z to the 7th power. Ans. —128 xyz", 
. Raise — 3 ab*c*u’e to the 6th power. 
Ans. 729 ab*cld'e°, 
. Raise ab to the 23d power. Ans. ab”, 
. Raise — a’b to the 23d power. Ans. — ab”. 
. Raise mn’c to the 8th power. Ans. m'n'®e°, 
. Raise — axz"y'™ to the 9th power. Ans, —a®a”y™, 
. Raise —10 aay" to the 10th power. 


Ans. 10,000,000,000 a1, 


. Raise 4 c'n™y? to the 5th power. Ans. 1024 c*n®y”, 
. Find the 3d power of — 8 cy’. Ans, — 512 cy”. 
. Find the 2d power of — 8 cy’. Ans. 64 c%y". 

Raise 2 + y to the 2d, 3d, 4th, 5th, 6th, and 7th powers. 


eo + 82°y + 8ay’? + y* = the 3d power. 

att 4a%y + 6a’y’? + 4ry> + y* = the 4th power. 
a + Saty + 10a*y? + 1l0ey’ + baytt+y = 

| the 5th power. 
a + Gaby + 15 aty? + 20 a2°%y? + lda’y* + 62xy° 

+ y® = the 6th power. 
x’ Taby + Wary’? + 85 aty® + 35 a*y* + 21 xy? 

+ 7 ary® + y' = the 7th power. 


yyy ELEMENTARY ALGEBRA. 


22. Raise « — y to the 2d, 3d, 4th, 5th, 6th, and 7th powers. 
Ans, 2? —2xy + y’ = the 2d power. 
x* —3 ay + 38axy’? —y* = the 3d power. 
at —42°y + 62°y? —4ay’ + y* = the 4th power. 
aw —b ay + 10 2y? —102’y> + dry* —y = the 
5th power. 
a —6ay + 15 2ty? — 20 ay? + 15 x’*yt —6 27° 
+ y° = the 6th power. 
x? —Ta’y + 21 ay? — 35 aty® + 35 x8y* — 21 ay? 
+ 7xy° —y' = the 7th power. 
23. Raise 2ab + 3d? to the 3d power. 
Ans. 8 a*°b* + 36 a’b’d’ + 54 abd* + 27 d®. 
24. Raise 2ab — 3d? to the 3d power. 
Ans. 8 a*b? —36 a’b'd? + 54 abd* — 27 d*. 
25. Raise 5 ed + n to the 5th power. 
Ans. 3125 ed’ + 3125 ctd'n + 1250 cd'n? + 250 c’d’n® 
+ 25 edn*t + n’. 
26. Raise 5 cd —n to the 5th power. 
Ans. 3125 ed’ — 3125 ctd'n + 1250 cd*n? — 250 c’d?n* 
+ 25 ednt — n°. 
27. Raise x + y —z to the 3d power. 
Ans. 24+ 382°y —82%4 y4 382ay —38y2z2—24 
3x2’ + 3 yz”? —6 xyz. 
28. Raise a + 6 —ec + d to the 2d power. 
Ans. a + 2ab + 0? —2ac + 2ad + &—2be + 
2 bd —2ed + ad’. 
243 cx’ 


my” 


Scones 
29. Raise —- to the 5th power. Ans. 
my 


30. Raise —— to the 4th power. 


625 at — 2000 akr + 2400 a2x? — 1280 ax® + 256 xt 


Ans. “5 — Said + Mai’ — Bae? + 16a, 





ELEMENTARY ALGEBRA. 213 


SECTION: LVI. 
The Binomial Theorem. 


THE BrnoMIAL THEOREM gives a short way of writing 
down, without actual multiplication, the powers of any bi- 
nomial. Let us take + y and x —y as specimen bino- 
mials, and consider their powers, as found in Examples 21 
and 22 of the preceding section. 


First. As regards the Sieans in these powers. We ob- 
serve that in Example 21, when the connecting sign of the 
binomial is +, all the signs of the powers are +. In Ex- 
ample 22, when the connecting sign of the binomial is —, 
the even terms (the 2d, 4th, 6th, and so on) are —. 


Sreconp. As regards the Exponrents. The first term 
of the binomial, x, has its leading exponent in each power 
the same as the number of the power. Thus, the 2d power 
begins with z?; the 3d power, with 2°; the 4th power, with 
xz*; and so on. The powers of « then regularly decrease. 
Thus, in the 5th power of the binomial in both examples, we 
have first 2°, then x‘, then x’, then x”, and then 2, in the suc- 
cessive terms. The powers of y, on the contrary, increase. 
They begin in the second term with y, then we have y’, then 
y*, then y‘, and then y’, ending as high as 2 began. Leay- 
ing out the coefficients, the terms would stand 

75 + xty + - dra + rs -|- xy + a 
4 ago Ly +4- xy? ws xy + xy* ss y’. 

Tuirp. As regards the Corrricients. The first coeffi- 
cient in each of the powers may be omitted, as it is 1. In 
the 2d term, the coefficient is the same as the number of the 
power. In the 2d power it is 2; in the 3d power, 3; in the 
4th power, 4; and so on. This is the case both in Example 
21 and in Example 22. Now, it will be found that if we 
take any term, multiply its coefficient by the exponent of its 
leading letter, and divide by the number of the term, we shall 


214 ELEMENTARY ALGEBRA, 


obtain the coefficient of the next term. Thus, in Example 
22, take the 7th power of « —y. Its 2d term is 7 xy, the 
coefficient being 7. Multiply 7 by 6 (the exponent of the 
leading letter, x). The product is 42. Divide this by 2, 
as we are at the 2d term. The quotient, 21, is the next co- 
efficient. Continuing in the same way, we may find all the 
succeeding coefficients. 


PROBLEM 1. Raise a —n to the 6th power. 


SoLturion. Since the connecting sign is —, the even 
terms (2d, 4th, &c.) will be —. 

The first term will be a®, and the powers of a will decrease, 
thus : a&—- a+ adé— &+ v— a. 

Inserting now the increasing powers of n, we have: 

a— an+ aiv— ani+ a’nt— ani+ n'. 

As to the coefficients, that of the 2d term must be 6, giv- 
ing thus far: a® — Gain. 

Multiply this 6 by the exponent, 5, of the leading let- 
ter, a, and divide by 2, as we are at the 2d term. 6 x 5 
= 30. Dividing by 2, we have 15, giving thus far: 

a’ —6a'n + 15 a‘n?. 

Multiply this 15 by the exponent, 4, of a, and divide by 
3, giving 20. So we have thus far: 

a’ —6a'n + 15 a‘n? — 20 abn’. 

Multiply this 20 by 3, and divide by 4, since we are at 
the 4th term. We get 15. So we have thus far: 

a’ —6a'n + 15 atn? —20 a'n®? + 15a’nt. 

Multiply this 15 by the exponent, 2, of a, and divide by 
5. We get 6. So we have thus far: 

a’ —6ain + 15 atn? — 20 a'n? + 15 a’nt —6 ani. 

Multiply this 6 by the exponent of a (1 understood) and 
divide by 6, as we are at the 6th term. Weget1. This 
need not be set down as a coefficient, since it will be under- 
stood without actually being there. So we have finally: 


ELEMENTARY ALGEBRA. 215 


a’ —6a'n + 15 a‘n? — 20 a'n3 + 15 a’nt —6 an> + n° 

This is the 6th power of a —n. It may be observed also 
that the coefficients vary from the end just as from the be- 
ginning. So that when half of them are known, the others 
can be written down. 

This mode of finding the powers of a binomial was discovered by 
Sir Isaac Newton. He left no proof of its correctness. Whenever it. 
has been tested by actual multiplication, the results have been found 
to be correct. It has also been proved that it must be correct in all 


cases. The proof may be found in most treatises which include the 
more advanced and difficult parts of Algebra. 


PROBLEM 2. Raise «* + 2 y° to the 4th power. 


Souution. In this example, we must perform the same 
operations on the terms as if they were single letters whose 
coefficients and exponents were 1 understood. We have 
explained how to use the Binomial Theorem in raising a 
binomial with such simple terms to the 4th power. Thus, 
we may take a + 6, and find its 4th power. It will be 
found to be 

at + 40° + 6a’b? + 4ab? + Ot. 

Now, instead of a put x‘, and instead of 6 put 2’, en- 
closing them in parentheses, so as to show what operations 
are to be performed on them. We then have 

(at) + 4 (at)? (2y*) + 6 (at)? (2y*)? + 4) ZY?) 

uj act 2 Y)' 

It is now seen that to obtain the true first term, 2* is to 
be raised to the 4th power; to obtain the second term, x‘ is 
to be raised to the 3d power and the result multiplied by 
2 y° and by 4; to obtain the third term, 2* is to be raised 
to the 2d power, so is 2 y*, and the results are to be multi- 
plied together and by 6. So on with all the terms. Per- 
forming these operations, the result is : 

x 4+ Say + 2aey + 3822ty? + 16 y”. 


Thus, if each term of a binomial is not simply a single 


216 ELEMENTARY ALGEBRA. 


letter, the Binomial Theorem may be applied to it just as 
if each term were so; but the terms must be enclosed in paren- 
theses, and the operations which are indicated must be per- 
formed on them afterward. 


EXAMPLES. 


1. Raise x + a to the 5th power by the Binomial Theo- 
remo: | 

Ans. x + 52ta + 10 aa? + 10 xa? + 5 rat 4+ a’, 

2. Raise a — «x to the 6th power by the Binomial Theo- 


rem. 
Ans. a& —6a5x + 15 atx? — 20 a3 + 15 a?2* — 6 aa5 
+ 7°, 
3. Raise c —a to the 9th power by the Binomial Theo- 
rem. 


Ans. &€ —9 &a + 36 cla? — 84 8a? + 126 bat — 126 
ca + 84 ca’ — 386 ca + 9 ca® — a’. 
4. Raise a + n to the 10th power by the Binomial Theo- 
rem. 
Ans. a + 10 an + 45 abn? + 120 an’? + 210 a®nt + 252 
an? + 210 atn® + 120 a'n™ + 45a7n® + 10 an? + n”™. 
5. Raise m —n to the 7th power by the Binomial Theo- 
rem. 
Ans. m’ —Tm*n + 21 min? —35 mn? + 35 mint —21 
mn + 7 mn? —n’. 
6. Raise 8 x —6 y to the 3d power by the Binomial Theo- 
rem. Ans. 27 28 —1622’y + 324 xy? — 216 y'. 
7. Raise 2? —y’ to the 8th power by the Binomial Theo- 
rem. 
Ans, x —8 x'ty? + 28 c?yt — 56 xy + 70 zy? — 56 
ey + 28 xty? —8 x7y"* + "6, 
8. Raise 5 « — y to the 5th power by the Binomial Theo- 


rem. 
Ans. 3125 a5 — 3125 «ty + 1250 ay? — 250 a’y? © 
+ 22ry* —y’. 


9. 
10. 


Ls. 


ELEMENTARY ALGEBRA. 217 


Raise « + ¢ to the 2d power by the Binomial Theo- 
rem. Ans. x? + 22e + ¢. 
Raise a + 6 to the 3d power by the Binomial Theo- 
rem. Ans. a + 3076 + 3 ab’? + 5B. 
Raise 52 + 3y to the 6th power by the Binomial 
Theorem. 


Ans. 15625 a® + 56250 ay + 84875 xy’? + 67500 zy’ 


12. 


13. 


14. 


16. 


ive 


18. 


+ 80375 2’y* + 7290 xy® + 729 y’. 
Raise 2m’? + 5n‘* to the 4th power by the Binomial 
Theorem. 
Ans. 16m” + 160 mnt + 600 mén® + 1000 min?? + 
625 n’*. 
Raise 5 p> —2q*° to the 4th power by the Binomial 
Theorem. 
Ans. 625 p® — 1000 p*q* + 600 pq* — 160 p'q° 
+ 16q”. 
Raise 2 ct —3m’n to the 5th power by the Binomial 
Theorem. 
Ans. 82c” — 240 c®m’n + 720 c?m*tn? — 1080 c8m*n? 
+ 810 ctm®n*t — 243 mn'. 


. Raise 3 m’n —2c* to the 5th power by the Binomial 


Theorem. 
Ans. 243 mn — 810 ctm®nt + 1080 c&mén3 
— 720 c?’m‘n? + 240 cm’n — 32 &”. 
Raise 1 —v to the 9th power by the Binomial Theo- 
rem. 
Ans. 1 —9v + 86v? — 840° + 126 0 — 1260 
+ 84 v§ — 36 0’ + 9v* —v°, 
Raise 2 « —1 to the 6th power by the Binomial Theo- 
rem. 


Ans. 64 «®§ —192 2 + 240 xt —160 2* + 60 2? 


—122 + 1, 
Raise 6 + 1 to the 3d power by the Binomial Theo- 
rem. Ans. 216 + 108 + 18 + 1. 


19 


218 ELEMENTARY ALGEBRA. 


19. Raise 3 — 3 to the 5th power by the Binomial Theo- 


1 (5 . 10. 10) See 
rem. Ans. Sag + gy > op 
20. Raise 5 = ; to the 4th power by the Binomial Theo- 
yi x ns ad (epee Z 2xn3 n* 
rem. rvs. 16 6 The “97, a1 


Notre. After performing the 18th example by the Binomial Theo- 
rem, add all the terms together and see if they are equal to the third 
power of 7. They should be so, since 6 + 1 is the same as 7. 


SECTION LVII. 
Evolution. Roots of Monomials. 


Evolution is the extraction of roots. Hence its processes 
are the reverse of the processes of Involution. 


Roots of Monomials. 


As the 2d power of a’ is a®, the square root of a® is a’. 
As the 8d power of a? is a®, the cube root of a is a’, And 
in all cases, as we multiply the index of a monomial by the 
number expressing the power to which we wish to raise it, 
so we must divide by the number expressing the root which 
we wish to extract. Thus the fourth root of a” is a’. 

To extract the cube root of a monomial composed of sev- 
eral factors, such as 64 #®y'®z”, is the same as to extract the 
roots of its factors separately. The cube root of 64 is 4, the 
cube root of x* is x’,and so on. The result, the cube root 
of the whole monomial. is 4 x°y5z’. 


oon oar WN & 


— he 
wo © 


fm 
OO 


14. 


15. 


16. 


17. 


18. 


19. 


20. 


ELEMENTARY ALGEBRA. 219 


EXAMPLES. 


. Extract the square root of 4 a’z’ 

. Extract the cube root of a®x"y’. 

. Extract the 4th root of 16 many’. 
. Extract the 5th root of 32 cd®e” 

. Extract the 3d root of 8 m<%d™. 

. Extract the 8th root of z*y**z®. 

. Extract the 10th root of np! 

. Extract the 3d root of 27 din 

. Extract the 4th root of 81 «*y"z*, 
. Extract the 5th root of 248 a7y'5z™. 
. Find the 3d root of 125 2®y*. 

. Find the 2d root of ne ral Te 


. Find the 2d root A ie ae 





b 
By 16 
Find the 4th root of ns 
Find the 3d root of Hee Zu; 
ah 
Find the 5th root of —— ay ye 


Find the 10th root of ~ a 
x 


: : 1 
Find the 6th root of 64 ty” 


Find the 3d root of 





27 
i 
Find the 2d root of = ays 


Ans. 2 ax. 
Ans, a’xty. 
Ans. 2 m'xy’, 
Ans. 2 cde’. 
Ans. 2 mic*d***, 
Ans. x*y*z°, 
Ans. np. 
Ans. 3 c’d’m® 
Ans. 3 xy*z. 
Ans. 3 27782", 
Ans. 5 x®y", 
Ans. 10 x*n”™. 


220 ELEMENTARY ALGEBRA. 


SECTION LVIII. 


Square Roots of Polynomials. 


To take the square root of x? + 2ay + y’, we must first 
find the square root of a part of it, for instance, of «. The 
square root of x’ is x. This is the first part of the root. 

We must next consider 2xy + y’, the rest of the quan- 
tity whose square root we are finding. If we double the 
first part of the root, obtaining 2, and divide the first term 
of 2xy + y’ by it, we get y, the next term of the root. 

This may be proved true by actual multiplication: «x + y 
multiplied by z + y will give 2 + 2ay + y’. 

We must observe that this square consists of two parts, 
x? and 2xy + y’. The latter part may be obtained thus: 
To the 2 a, the partial divisor by which we got the term y, 
add that term, making 2x + y. Multiply the divisor thus 
completed, by its last term, y, and we obtain 2 xy + y’. 

Now what we have said is appiicable to any quantity 
which has a square root. The first part of the root may be 
operated on as x was above, and the next term found as y 
was. This must be true; for x and y may stand for the two 
parts of any square root, the square of which must then be 
the same as x? + 2ay 4+ y’. 


Before extracting the square root of any quantity, its terms 
should be arranged as in Division. 


PROBLEM. Extract the square root of 4a* —12a°5y + 
29 aty? — 30 ay? + 25 ay. 
OPERATION. 
Root. 
4 x6 —12 x5y 4 29 xty2 — 80 x3y3 + 25 atyt ( 223 —3 xy + 5zxy? 
4 x6 
4733 a) —12 ay + 29 xty? — 30 x3y3 4 25 x2y4 
—12ay+ 9xty? 


4a —6 ay + d xy?\ 20 xty? — 30 x3y3 4 25 x24 
20 xty? — 80 x3y8 4. 25 x2y4 








— 





ELEMENTARY ALGEBRA. 921 


EXPLANATION. Here the terms are already arranged 
according to the powers of x, beginning with the 6th, then 
the 5th, then the 4th, and so on. 

The square root of 42° is 2%. We set down that much 
of the root, and subtract the square 4° from the whole 
power. 

We then double 2 2’, giving 4°, and divide that into the 
first term of the remainder. It gives —38a’y. We set 
down —3zx’y in the root, and also after the partial divisor 
42°, and multiply the divisor thus completed by —32’y. 
It gives —12a°y + 9 x*y’, which we subtract from the part 
of the power which we are considering. 

We then double 2 x? —3 2’y, the part of the root already 
found, and set down as before the product, 4 2* —6 2x’y, as 
a partial divisor. Dividing, we get + 52y’, which we set 
down in the root and after the partial divisor. Multiplying 
the completed divisor by 5 xy’ and subtracting it from the 
remainder of the power, we find that nothing remains. 


EXAMPLES. 


Find the square roots of the following quantities. 


1a2+4ay+4y’. Ans. x + 2y. 
Daa + Jake -- a%s'. Ans. a? + ax. 
3. a + Qaix -- x’. Ans. a® + 2, 
4. at —6a*xr + 9 2’. Ans. a — 382. 
5. a®a? —2 a's’ + ax’. Ans. atx —ax*. 
6. ct —10c’d + 25 d’. Ans. & —5d. 
7 1+10c + 25e’. Ans. 1+ 5e. 
8. 225 + 90m + 9m’. Ans. 15 + 38m. 
9. 36 +1241. Ans. 6 +1 = 7. 

10.27 + 2ay+2az+yY+2yz4 2. 
Ans. «© +y +4 


19 * 


222 ELEMENTARY ALGEBRA. 


11. 4a* + 20 ae —4am + 256 —10em + m. 
Ans. 2a + 5¢—m 
12.14+ 684+ 2:4+9%+68st+8. Ans. 1+3848 
13.9+ 6m+ 48n + m’? + 16mn + 64n7. 
Anz. 3+ m+ 8n. 
1441+4+104+ 1249. Ani 14+2+3=6. 
15. 4074+ 44+ 4ee¢* +14 22 4+ x. 
‘ Ans. 2a +1 + 2%. 
16. 4é—4é +46é—34é—2é+ 3¢€—2et lL. 
Ans. 246—e +e—1. 
17. a2 + 2ab + 2ace—2ad + & + 2he —2dbd +e 
—2ed + d’. Ans. a+6+e¢e—d. 
18. 2° + 227’ —2*§ + Bet — 224+ 2, 
Ans. o*+ FP —#+4+ 2. 
19. 252° + 202 + 142+*—6e°—3xr—22r+1. 
Ans. 52° + 227 + « —1. 
20. 49 2*—1l4axr* + 15 es? —2e2 + at. 
Ans. 7 x? —azx + at 


SECTION LIX. 
Classification of Equations. 


Simple equations are those in which the unknown quanti- 
ties are all of the first power. Simple equations are also 
called equations of the first degree. The equations thus far 
given in this book have been of the first degree. 

Quadratic equations, or equations of the second degree, are 
those which contain the second power or the second root of 
an unknown quantity, or the product of two unknown quan- 
tities; but do not contain any higher power or root, ora 


ELEMENTARY ALGEBRA. 223 


product of more than two unknown quantities. They may 
also contain the first power of an unknown quantity. If 
they do, they are called complete or adfected quadratic equa- 
tions. If they do not, they are called incomplete or pure 
quadratic equations. 


Cubie equations contain the third power or root of an un- 
known quantity, or the product of three unknown factors; 
but no higher power or root, or product of more than three 
unknown quantities. Cubic equations are of the third degree. 


Biquadratie equations contain the fourth power or root of 
an unknown quantity, or the product of four unknown fac- 
tors; but no higher power, root, or product. They are of 
the fourth degree. ; 

Equations containing higher powers, roots, or products, 
are of the fifth, sixth, seventh, &c., degrees. 


A simple equation, x—6 = Tx —50 —2z. 
x? = 36 
Pure quadratic equations, nn estye 
xy = 48 
x? + xy = 52 
( om Am =i 
Adfected quadratic equations, | os i fe a a0 
Liy + %at = 408. 
A eubic equation, et 24+ x= 39. 
A biquadratie equation, z*—5aS+ 3 = 64. 
An equation of the fifth degree, a es eee 


An equation of the sixth degree, 
Fa? —32' + ot—47 —2' + 2—4= 1, 





224 ELEMENTARY ALGEBRA. 


SECTION LX. 


Pure Quadratic Equations, and others Similarly 
Solved. 


PROBLEM 1. Given x’? = 49, to find the values of a. 


SotutTion. Any square number, such as 49, has two 
square roots; for, + 7 multiplied by + 7 gives 49, and 
— 7 multiplied by —7 gives 49. Consequently, the value 
of x may be either + 7 or —7, not both at once, but either 
one. The operation is generally set down thus: 

a? = 49. 
The double sign should be read plus or minus. 


ProsLEeM 2. Given (x + 3)? = 6a + 34, to find the 
values of x. 


Soxtution. This is not strictly a pure quadratic equa- 
tion, as it contains the first power of x. But, as will be 
seen, it can be easily reduced to one. 

(2 + 8)? = O@ + 34. 
Squaring thew +3, 27+ 6x2+9= 62 +4 34. 


Cancelling, x? +9 = 34. 

Transposing, x? = 84 —9. 
Chee 

Taking the square root, est 8. 


Hence x may either equal + 5 or —5 in the given equa- 
tion. By substituting either + 5 or — 5 for # in that 
equation, the results will be found to be equal, thus proving 
the answer. 

The true value of an unknown quantity, when substituted 
in an equation for that quantity, is said to satisfy the equa- 
tion. That is, the equality is preserved. 


ELEMENTARY ALGEBRA. 225 


: a 
PROBLEM 3. Given 5 — a = le =; aS : + - to find 


the values of 2. 


Sotution. As this stands, it is a simple equation. By 
clearing of fractions, it becomes a pure quadratic equation. 


ee a [We ae Saa eee 

GAS 36 mem Gre OF 
Clearing of fractions, 4%? — 2? = 2592 — 9x’ + 6x* + 42x’. 
Transposing, Ag? — x? + 9x? — 6a* — 42? = 2592. 
2a* == 2592. 
we == 1200: 


Propiem 4. Given /9 4+ 2? = 5, to find the value of x. 


SoLuTION. J/9 + x2? = 5. 
Squaring both members, 9 + a2? = 25. 
a = 25 —9. 
a ea 16: 
x= +4, 


It must be known that to square the square root of a 
quantity is to obtain the quantity itself. Hence, to square 
a radical quantity, remove the radical sign, provided it is 
the square root which is indicated by that sign. 


/104—a’ —2x 


PropiEeM 5. Given a 4, to find the 
values of a. 
S V¥ 104 — 2’ — 2 
OLUTION. *¢ceeier srapmmpr re ra 4, 
Clearing of fractions, V¥104 — 2’? —ax = 4a. 


It will not do to square yet, as the radical quantity does 
not stand by itself. 


P 


226 ELEMENTARY ALGEBRA. 


Transposing, Vv 104 —a” = 4x + ov, 
/104 —a? = 5a. 
Squaring now, 104 —x? = 252’. 
Transposing, — 25x? — x2? = — 104. 
Changing all the signs, 25x? + a? = 104. 
26.4 =i: 
Ere Pf 
EV he 


Propiem 6. Given Y 4 + 2V¥ 160 — 82 = x — 2, to 


find the values of x. 


Sonution. “4+ 2160 — 8a = x —2. 
Squaring, 4+ «2/160 — 8x" = a? —4r + 4. 
Cancelling 4, x/160 —8xr = 2 —4z. 
Dividing by 2, V160 — 8% = « —4, 
Squaring again, 160 — 8x = x’ —8xr-+ 16. 
Cancelling — 82, 160 = a? + 16. 

—x’ = + 16 —160. 
a? = —16 + 160. 


x? = 144. 
xt use 
EXAMPLES. 
1. Given «2? —15 = 49, to find the values of 2. 
Ans. c= + 8. 
2. Given 5x? + 5 = 50, to find the values of =. 
Ans. & = = 38. 
3. Given 8x? — 12 = 8 + 372’, to find the values of 7. 
Ans. @ == + 2. 
4, Given 7x? —90 + 8 = 45 —15, to find the values of 
be Ans. & = 4s 


5. Given 9x? —4 = 0, to find the values of z. 
Ans. & == re q. 


ELEMENTARY ALGEBRA. 227 


6. Given 3x? —72 = 22? + 72; to find the values of 2. 
Ans. x = + 12. 


7. Given 28x27? —7 = 28x? — 52, to find the values of z. 
Ans. « = +3. 
8. Given sain + islet = 12, to find the values of x 
2—x 242 ‘ k 
Ans. © = £1. 
9. Given 3a? —4 = 121 — 2z’, to find the values of za. 
Ans. c= t5. 
10. Given pie -- bode = 8 to find the values of x 
. 1l+2z2 -l1—c« 38 : 


Ans. a — i ze 
11. Given (x + 4,? = 8x +4 25, to find the values of x. 
Ans. © = + 8. 





2 
12. Given e x? — (52? —16) = es x uo to find the 
values of x. Ans. © = +5. 
: Bt 8x 
13. Given 3z Tea Se to find the values of z. 
Ans. x = + 9. 
14. Given (4c + 1)? = 2x + 1,4, to find the values of 2. 
Ans. © = t1 


4° 
; bai + 3° Zz? xe+2 
15. Given 9 ae ma = 4a* + 19” to find the values 


of 2. Ans. © = +3. 


627+ 3 
Note. First perform the subtraction re 3 pee Bee , and re- 


duce the resulting fraction to its lowest terms. 


A 1 3 108 
16. Given Dan? + 8 = gi 1 find the values of 2. 
Ans. Ci oo ys 
17. Given eae + surat B20 —, to find the values 
rej” gt} 
of 2. Ans. x = + 8. 


228 ELEMENTARY ALGEBRA. 


18. Given oe = 45, to find the values of z. 

Ans. © = + 6. 
19. Given “32? + 6 = 9, to find the values of z. 

Ans. % = + 5. 
20. Given “a? + 27 = 2,/9, to find the values of a. 

Ans. «x = + 8. 
21. Given x + 7x = 22%, to find the values of 2. 

Ans. 2 = + 2, 
22. Given “2? + 15 —5 = 3, to find the values of 2. 

Ans. x = + 7, 


In the following examples, the equations, when reduced, 
are not quadratic, and x has but one value. 


93. Given “3x + 10 = 5, to find the value of z. 





Ans. x = 5. 
24. Given V 21+ x = 7 — // 2, to find the value of z. 
Ans. % = 4, 
25. Given Vx —5 = ./x —1, to find the value of z. 
Ans. «© = 9. 
26. Given Va —11 = /x + 1, to find the value of 2. 
Ans. x = 36. 
27. Given Vx —9 = ./x —1, to find the value of z. 
; Ans. x = 25, 
28. Given /3x —11 = 3x —1, to find the value of z. 
Ans. x = 12. 
29. Given “4 +2 Va? —80 = « —2, to find the value 
of x. Ans. x = 12. 


30. Given « + VW 16x + a? = 8, to find the value of z. 
Ans. x = 2. 





ELEMENTARY ALGEBRA, 229 


SECTION LXI. 
Equations with Two Unknown Quantities. 


WHEN pure quadratic equations contain two unknown 
quantities, various artifices are employed for elimination, 
the most general of which will appear from the following 
examples : 


Propiem 1. Given 2” ey 3 i} to find the values 


of x and y. 
SoLution. Squaring the members of the first equation. 


e+ dey+y’ = i. 
Subtracting 4 times the 








members of the second, Ary = 56. 
x* —Qey + y? = 25 
Taking the square root, xe—y = +5. 
Adding the members of 
the first equation, e+ ys 9. 
22 = 14 or4 
x eax OF aw 


Substituting the values of x in the 
equation, 7 + y = 9, we have y = 20Y /. 
Wrosec that if a2 = 7, y = 2; but, if 7 = 2, y =. 7. 


Prosiem 2. Given ‘a BE de ul } to find the values 


L—y = 
of x and y. 
Soxtution. The first equation is, 
oe + y? = 65. 
Squaring both members of the 
second, a? —Qry + y? = 49. 
Subtracting, Qxey ‘ox 16, 
Adding the members of this to 
the members of the first 
equation, a? + Qey+ yy’? = 8l. 
Extracting the square root, at-y = +9. 


20 


230 ELEMENTARY ALGEBRA. 


With this equation and the second of the original equa- 
tions, the values of x and y may be found as in the preced- 
ing problem. It will be found that z may be equal to 8, in 
which case y = 1, or x may be equal to — 1, in which case 
y= —8. 


| EXAMPLES. 
1. Given ‘i La =e . to find the values of # and y. 
Ans. x = 7 or —1; y = 1 or — 7. 
2. Given ie ate cs on to find the values of x and y. 
Ans. «x = 50r3;y = 8 or 5. 
3. Given Fay a " rs ibe to find the values of x and y. 
ge ke Ans. « = 9 or 3; y = 2 or 36. 


Norse. This is similar to on two preceding it. 


4, Given se re ge a to find the values of x and y. 


Ans. « = 4or —2; y = 8 or —6. 


Norts. This also is similar to the preceding, only it is not 4zy that 
is to be added to the middle term after squaring the members of the 
first equation, but 24zy, as that is double the middle term. 


Rips, Sao) 

5. Given iy «x "+ to find the values of # and y. 
v+ y= 82. 

Ans, © = 2243 y = 4, 


Nore. Clear of fractions, and then find what 2zy equals. 


6. Given 13, ihe is 1 to find the values of x and y. 


Ans. «== 19 See 


a+ y’ = 10. 


gah y meee. t to find the values of x and y. 


7. Given | 
Ans. & = 380r1; y=1lor3. 


ELEMENTARY ALGEBRA. 231 


8. Given he ms = 240.) 14 find the values of 2 and y. 


tr me $25 
Ans. © ese 435 y= & 2, 


Note. First divide the members of the first equation by those of 
the second. 


9. Given 1 =| sid 3. to find the values of x and y. 
| Ana 2226: Y= 2. 

: c —y’ gs WA 
10. Given | oy 8. to find the values of x and y. 
Anas 2 22:6) .y =2'7. 


SECTION LXII. 
Adfected Quadratie Equations. 


IF we square a monomial, the result isa monomial. Thus, 
if we square 7a’x*, we get 49a‘tzx”. 

If we square a binomial, the result is a trinomial. Thus, 
if we square x + a, we get az? + 2ax + a’. 

Reversing these operations, we see that we can take the 
square root of a monomial or of a trinomial. We cannot, 
however, take the square root of a binomial. 

Let us consider the equation 

w -+- 62 = 16. 
We cannot take the square root of x’ + 62, since it has but 
two terms; but we can add something to it which will make 
it a complete square, and then take the square root. By 
Theorem I., page 96, the square of the sum of two quantities 
as equal to the square of the first, plus twice the product of the 
first and second, plus the square of the second. Now here we 
may consider x? the square of the first, and 6x twice the 
product of the first and second. Then 3z will be once the 
product of the first and second. Dividing by x, which is 


932 ELEMENTARY ALGEBRA. 


the first, we find that 3 must be the second. The square of 
this is 9, which must be added to x? + 6z, and also to 16, 
to preserve the equality. 7 
a? + 6x = 16. 


9= 9. 
xz’? + 62 + 9 = 25. 
Taking the square root, £ +o eee 
c= +5 —8, 
x = 2 or —8. 


The operation we have just explained is called completing 
the square. 


Prospiem. Find the value of x in the equation 
a+ x—l1 3 
re eel a 


So.tution. This equation is not quadratic as it stands, 
but becomes so when cleared of fractions. 

Clearing of fractions, . 
227 + 4a + 2 = 2x? —4a + 2 + 382? — 8. 

Collecting, — 3x7 + 8x4 = — 3. 

Changing signs, 382° — 8& = 3. 

Now, as the first term is not a perfect square, it must be 

made so. This can always be done by dividing by the coef- 

ficient of x’. 

Performing this division, go — ba = I, 

In order to complete the square, consider that 2 must be 
the first term of the root, and —8z twice the product of the 
two terms. Then — 4x must be once that product. Divid- 
ing by x, which is the first, we find that — #4 must be the 
second. The square of this is 4°, which must be added to 
x” — 8x, and also to 1, to preserve the equality. 


x ee 8a =_ i 
6 





Taking the square root, 


ELEMENTARY ALGEBRA. 2aG 


emts + FZ. 
xv = 3 or —}#. 


These operations should be proved by substituting each of the 
values of x separately in the original. equation. 


EXAMPLES. 


Find the values of x in the following equations: 


10. 
th 
12. 


13. 


15. 
16. 
17. 


18. 
19. 


. 2+ 8r = 20. 
. a —2r = 3. 
a+ 4a = 45. 
ee — 4x = 12. 
x? —10x = 24. 
mae 2 = 90. 
a+ a= 10 
“+e = 3 
.a—x=6 
xz’ — 102 = 56 
xe? + Tx = 60. 
x? — bx = 24. 
xr+l x—d 
peeeey 3°10 
x—3 2x£—24 
dy. = 7800 
ha? — 3x0 + T = 42? + 5. 
3x0? —9x = 12. 
x? —13x2 = — 40. 
x4+2 8 
iegecinmet 
4x? —9a = 385. 
ers ig, 
2 


20. 


1 
2 
3 
4 
5. 
6 
(| 
8 
9 











20 * 


Ans. x« = 2 or —10. 
Ans. x = 3 or —1. 
Ans. x = 5 or —9. 
Ans. x = 6 or —2. 

Ans. « = 12 or —2. 

Ans. x = 9 or —10. 
Ans. x = % or —8. 
Ans. x = 4 or —3. 
Ans. x« = 3 or —2. 

Ans. x = 14 or —4. 

Ans. « = 5 or —12. 
Ans. x = 8 or —3. 


Ans. x = 20 or 1. 


Ans. « = 18 or 2. 


Ans. x« = 2 orl. 
Ans. x = 4or—l. 
Ans. x = 8 of 5. 


Ans. x = 2 or — 5h. 
Ans. x = 11 or — ¥. 


Ans. « = 5or0, 


234 ELEMENTARY ALGEBRA. 


Nore. In eee 20, if the value 0 be substituted for x, we shall 


have e 
0+0 
oT =0+6. 


This reduces to . ~ = 6. 


This equation cannot be said to be false, as 6 times 0 are equal to 
0. It is, however, meaningless, as far as the science of Algebra is 
explained in this book. In more advanced treatises, a meaning is 
found for it, and the fraction $ is said to have any value whatever, 
since any number of times 0 are equal to 0. 


SECTION LAXIII. 
Miscellaneous Examples. Review Questions. 


1. Drv1pE a line 54 inches long into two parts differing 
by 14 inches. Ans. The parts are 20 inches and 34 inches. 


2. Divide $118 among A, B, and C, so that B shall have 
$i2 more than A, and C $10 more than B, Ans. A must 
have $28; B, $40; and C, $50. 


3. A laborer was engaged for 25 days. For each day he 
worked, he was to receive $1.00 and his board. For each 
day he was idle, 50 cents were to be deducted for his board. 
At the end of the 25 days, he received $17.50. How many 
days did he work, and how many days was he idle? Ans. 
He worked 20 days, and was idle 5. 


4, A hare is 25 of her own leaps in advance of a grey- 
hound, and takes 11 leaps while the greyhound takes 10. 
Five of the greyhound’s leaps are equal to six of the hare’s. 
How many leaps must the greyhound make to overtake the 
hare? Ans. 250. 


Notre. If 5 of the greyhound’s leaps are equal to 6 of the hare’s, 


ELEMENTARY ALGEBRA. 235 


10 of the greyhound’s must be equal to 12 of the hare’s. Thus the 
greyhound gains one of the hare’s leaps while taking 10 of its own. 


5. Divide 45 into 3 parts, so that the second may be three 
times, and the third five times the first. Ans. 5,15, and 25. 


6. Divide 45 into 3 parts, so that the second may be 
double the first, and the third three times the second. Ans. 
5, 10, and 380. 


7. A’s age is half of B’s. B’s is three times C’s. The 
sum of their ages is 110 years. What is the age of each? 
Ans. A is 30 years old; B, 60; and C, 20. 


8. A, B, C, and D have $100,000. B has twice as much 
as D. C has as much as A and D together. <A has 2 as 
much as B. How much has each? Ans. A has $20,000; 
B, $30,000 ; C, 35,000; and D, 15,000. 


9. A, B, and C together have $3600. If C give B $300, 
B will have $100 more than C. If B give A $40, A and B 
will have the same amount. How much has each? Ans. 
A has $980; B, $1060; and C, $1560. 


10. If B give A $4000, A will have six times as much as 
B will have left. If C give B $2000, B will have four 
times as much as C will have left. If A give C $6000, C 
will have five times as much as A will have left. How 
much has each? Ans. A has $8000; B, $6000; and C, 
$4000. 


11. Find a number represented by three digits whose sum 
is 8, the digit in the place of tens being double that in the 
place of hundreds, and the order of the digits being inverted 
by adding 396 to the number. Ans. 125. 


12. A, B, and C have $4400. If B give A $1000, A will 
have $290 more than C. If A give B $90, B will have § 
as much as © has. How much has each? Ans. A has 
$1090; B, $1510; and C, $1800. 


236 ELEMENTARY ALGEBRA. 


13. Divide $1400 among A, B, and C, so that $400 + 10 
times A’s share, $240 + 6 times B’s share, and $720 + 
twice C’s share, may be equal to the same sum. Ans. A’s 
share is $200; B’s, $860; and C’s, $840. 


14, What fraction will equal 2 if 6 be added to its numer- 
ator, and 4 if 3 be added to its denominator? Ans. +4. 


15. What fraction will equal 4 if the numerator be 
doubled and the denominator increased by 5, and will equal 
4 if the denominator be doubled and the numerator in- 
creased by 5? Ans. 3. 


16. A wine-merchant mixes wine which cost him $20 per 
dozen with wine which cost him $25 per dozen, so as to 
make in all 36 dozen, costing him $212 per dozen. How 
many dozen of each kind did he take? Ans. 24 dozen at 
$20, and 12 dozen at $25. 


17. I have three weights. The first is 73 lbs. The 
second is as much as the first and half the third. The third 
is as much as the first and second. To what do they all 
together amount? Ans. 60 lbs. 


18. If A and B should give C ten sheep each, C would 
have 38 more than they would have together. If A and C 
should give B ten each, B would have 16 more than they 
would have together. If B and C should give A ten each, 
A would have just as many as both the others. How many 
has each? Ans. A has 18 sheep; B, 21; and C, 32. 


19. Divide the number 225 into three parts, so that twice 
the first part diminished by 5, twice the second part in- 
creased by 15, and three times the third part diminished by 
70, may be equal. Ans. 80, 70, and 75. 


20. A number represented by three digits is 25 times 
their sum. The figure in the place of units is one more 
than the sum of the other two, which are equal. What is 
the number? Ans, 2265. 

20 * 


ELEMENTARY ALGEBRA. 937 


21. Find the values of a, y, z, and w, in the equations, 


Ans. = 3; y=2;2=1lsw= 0. 
Note. Since wz = 0, either w or z must be 0. It cannot be z, for 
then zz would be equal to 0. Therefore w = 0. 
Taking the first, second, and fourth equations, find the values of 
“, y, and z. 


22. Find the values of # and y in the equations, 


egies 

a y \ 30 

ita Bune) 

a visiap re BO! 
Anse t= 0 31- eee; 

28. Add Ato of SM to } of 3 of §. 

laa 
Ans 0st 


19a7e*x? Amix? 








24. Multiply nin PY Beare’ add 1 to the product, and 
reduce the resulting mixed quantity to the form of a fraction. 
age tloge 
aera Ee TE 
25. Change 18acx to a fraction whose denominator shall 
be dmn‘z’, 90aemnix* 
Del See 

omnia? 





: 21a* : 
26. Change —— to an equivalent fraction with 6mn’x 


for a denominator. 126a3 ca 
A 
6mn2x 
27. Change a + c to the form of a fraction having a —e 
for a denominator. Ae. a —¢ 





a—c¢ 


238 ELEMENTARY ALGEBRA. 


28. Divide —lLby@+e¢+e6+e+e41. 
Ans. e —1. 
27a’c —27Tex? 


26 atot — Bocigt 0 US lowest terms. 


29. Reduce 











Ans. Te + doa 
1 1 a 
30. Subtract aay from eaaays Ans. aye 
——0bg400—. 


REVIEW QUESTIONS. 


Wuat is a monomial? Which of the following quantities are 
monomials, and which are polynomials ? 


a*be — abc?, 
36a — 868, 
36.062, 
a+b+e, 
abe, 
a? +. 2ab + 52, 
z+ 2y—4z—w-+n, 
atyzwnp. 


In the polynomials above given, are there any binomials? any tri- 
nomials? if any, name them. 

How many terms are there in the above expressions ? 

Which of them have coefficients expressed? Name the coefficients. 

What are the values of the following expressions ? 


a/9, 8/27, 4/16, 74, </8; 4/25, fee 


What is the difference between (6 + 4) x 5 —38 and (6 + 4) x 
(8)? 

Are the terms zy and zy? similar? Are 36z and 12z similar? 
Are 7x’y and 136z*y similar? 

Explain the subtraction of —3a from 7a; of 4a from —6q; of 
— 4a from —6a; of a from 56; of —a from 8. 

What is the rule for signs in subtraction? in multiplication and 
division ? 


ELEMENTARY ALGEBRA. 239 


What is the difference between arithmetical addition and algebraic’ 
addition ? 

State all the ways in which multiplication may be indicated. 

What is meant by transposition? Why must the sign of the trans- 
posed quantity be changed ? 

What is the rule for the indices in multiplication? in division ? 

In multiplication or division of polynomials, how should the terms 
be arranged ? 

What is an exponent? Give an example. 

What is meant by elimination? What are the three methods of 
elimination in common use? 

How many equations are necessary for the solution of any alge- 
braic problem involving unknown quantities ? 

What is a factor of a quantity? a measure of a quantity? Give 
four important theorems useful in factoring. 

What is a prime quantity? a multiple of a quantity? 

What must be done to two fractions of unequal denominators in 
order to add them ? 

How may an equation be cleared of fractions ? . On what principle 
does the operation depend? 

What is a ratio? a proportion? What are the terms of a ratio 
called? Which terms of a proportion are called extremes? which 
are called means? 

What is meant by a mean proportional between two terms? 

Prove that the product of the extremes of a proportion is equal to 
the product of the means. 

What is an arithmetical progression? its common difference ? 
Give a formula for finding its last term, and one for finding the sum 
of the terms. Prove them. 

What is a geometrical progression? its ratio? Give and prove 
formulas for finding its last term, and the sum of the terms. 

What is involution? evolution? Which powers of a negative term 
are negative ? 

Explain the process of finding the square root of a polynomial. 

Explain the binomial theorem. 

Why may we change the signs of all the terms of an equation? 


it 


240 ELEMENTARY ALGEBRA. 


SECTION LXIV. 
Negative and Fractional Exponents. 
By the principles of division, 
a--a=a’, 


av~—a=a. 
a+a=l1., 


a | pate cree 

a 
die: 5 ea 
PR RP dal OY 
1 1 
aha: 


In the first of the above operations, the exponent of the 
divisor is subtracted from the exponent of the dividend to 
give the exponent of the quotient (1 understood subtracted 
from 3 gives 2). If we do this through them all, we shall 
have 


a —-a=a 
e + a4 
a ~+-a=a 
a +a=a-}, 
a ++~a=a~?, 
a-~*~a=a~’, 


Comparing the two sets of operations, we may say a° = 1, 
1 1 


Oise a7~* = oe? and a~*=-—. Hence, though a* means 
a 


the product obtained by taking a as a factor 3 times, we see 
that a—* does not mean the product obtained by taking a as a - 
factor —3 times (which is nonsense); but a—* means 1 _ 
divided by a*. A quantity having a negative exponent means 
unity divided by the same quantity with a positive exponent. 
That a? = 1, may be proved by dividing any power of a 


ELEMENTARY ALGEBRA. 241 


by itself. Thus, a°-+a'°=a°=1. Anything divided by 
itself gives 1. So m‘n?y + min’y = mn" = 1. 
By the principles of evolution, 


J/al? = a’, 
AAO = Oe 
/a = a. 


In the above operations, to take the root of a quantity, we 
divide its exponent by the number which indicates the de- 
gree of the root. To take the cube root of a”, we divide 12 
by 3, giving 4 for the exponent of a in the root. If we do 
this even when the division gives a fractional result, we shall 
have cases of the following kind: 


ree: 
i/a= a. 
Fhe or 


: : 
Here we see that c° does not mean the product obtained 


by taking ¢ as a factor ? times (which is nonsense); but e 
means the dth root of c’. The denonunator of a fractional 
exponent indicates the degree of a root to be taken; the numer- 
ator indicates the power. 


EXAMPLES. 

1. Divide 2 by 2. Ans. 2 = 1, 
2. Express x—*y—* without negative indices. Ans. =; 
oe Pek 

3. Express “ip ine form which is not a fraction. 
3 Ans, 2am 


4, Express ; in a form which is not a fraction. 


Ans, Cie 
21 


242 
5. 


10. 


HE 


12. 


13. 


14, 
15. 


. Divide x’ by 2’, expressing the result by 2 with an in- 


ELEMENTARY ALGEBRA. 


Express a*y’z—* without a negative index. 
Z 


dex. Ans. x~5, 


. Add 2—?, s 5277, —T2-?, —5, and 827", 


5 
= 9 a 
Ans. 5a or > 


9 bined alt 5 Scie 11 





. Add 8z7*a—*, -— —~, 9=,-, and 
a x akx 


0 
. Add 15a°x-5, oe shee ——, and #—*, 
a? 2 a 


Ans. “= 242-, 
From 2a’n—'— y~*ax + 1 take 1 — 9axy—? — 3n~'a’. 


Ans. Bann! 4 Bary? = 
n y 
From 162—?— 5y~? take 152-7 + > 
‘ 2 ol ees 
Ans. x~?—8y Bea 
Ta 5a 
—3 —4 MEADS, Siem ee 
From az—* + ay~‘ take Sorry 
6a , 6a 
a =8 obese Sly 
Ans. —6aa—* + 6ay = + 7 


From 162~*y—* take i iiss 


z rag) 3 
16 
Ans. 16a-*y—* — 5abys = pe 
Multiply 25 wWe&y’x by — 3 a*zy°ex. Ans. — Th wv exy’. 


Multiply a—* by a-*. Ans, a~* = as 


16. 


17. 
18. 


19. 
20. 
21. 


22. 


23. 
24. 


25. 


26. 


27. 


28. 


ELEMENTARY ALGEBRA. 243 


Multiply 5 a%x—*— 3 a’e—*a? + 15 ¢%a* by 2a "a —*. 
Ans. 10a71cetx—* — 6 ace—24—* + 304a7—'e%2? 
_10¢ 6a , 30¢%" 


att eat us a 
Divide a~* by a~*. Ans. a. 
Divide c~*x—*y? by e’a—*y, Ang. 6 Ry == Z, 
Divide x” by x”, Ans, 2*—" = =e. 
Divide x” by x”. Ans, 2 =I, 


Add x?,—17 27, 9/2, 5x?, and 13,/z. 

Ans. 11 a? ae ae 
Add 13a*, —53/a,—T72/a, 214/a, and a’, 

Ans. 8a = 234/a, 


Add / az and ata. Ans. 2/ ax = Qatx?, 
Add /az and phe Ans. Jax + on 


From 29 atxty? take 16 Jaixy. 
Ans. 13Jaay = 13a'x'y?, 
Multiply —17 ata? by — Qata8, 
Ans. 34a°x® = 34 aad 


Multiply at —atat ao ot by at + aot 
Ans. oe + ot = Ya + 2%, 
Divide ax? by —5a%x?, 
$ | Lr 
Ans. a a ‘a= — = —; a 
5a ° 


244 ELEMENTARY ALGEBRA. 


29, Divide —17a'x* by —2a%a?. 


17 3 ae 
Ans. aa = i ae 


80. Divide at —2* by at —2’. 
Ans. a+ atat + ata? + ata? 
at a ee 
Ans. 9a? — 30a'*x® 4+ Gata + 25 atat 
—10 ata + at. 


+ 2x. 


31. Square 3at —5aé 


od 
| 


—1l1 
32. Square ee 
a*"— 2x 
i 
"SO 9a? + at m (a —2 ax? + at) 
33. Find the sq. root of a7? + 2a—'b—* —2a7*e-* + 
b-§§—2b-%e-* + e-8, Ans. a-! + b-§—e-+, 
34. Raise — Bat bic? to the 2d, 3d, and 6th powers. 
Ans. 25ab%c®, —125a2b%, and 15625 a°b°ct 
35, Find the 2d and 3d roots of 64a7a—*. 


t 4 
Ans. Bata = oe and bgt = an 


SECTION LXV. 
Radicals. 


A radical expression is an expression containing a radical 


2 


sign or a fractional index. Thus, ./x and c* are quadratic 


radicals or radicals of the 2d degree; Wa* and 3a* are cubic 


ELEMENTARY ALGEBRA. 245 


radicals or radicals of the 3d degree; 4/y and an* are bi- 
quadratic radicals or radicals of the 4th degree. 

Any algebraic expression may be reduced to a radical 
form, as the following examples will show. 


PROBLEM 1. Reduce mn + y to a quadratic radical. 


Sotution. The square of mn + y is m’n? + 2mny + y. 
Hence the square root of mn’ + 2mny + y’ is mn + y. That 


is, mn + y=V mn? + 2mny + y’. 
PROBLEM 2. Express 3x+/ a—-x entirely under the radi- 


cal sign. 


Sotution. The 3d power of 3x2 is 27x%. Hence the 


given radical is the same as 7/272°x Ya—zx. Multiply- 
ing together 272* and a—z, we have 27az*— 272+. To 
take the cube root of this is the same as to take the cube 
roots of its factors and multiply them together. That is, 


3/27 ax */a—= is the same as 3/27 ax® — 27 x. 


GENERAL PRINCIPLE OF EyouutTion. To take any root 
of a quantity is the same as to take the same root of each of 
its factors and multiply the results together. 


EXAMPLES. 
1. Reduce 8 to a quadratic radical. Ans. / 64. 
2. Reduce 4 to a cubic radical. Ans. </64, 
3. Reduce 5 to a biquadratic radical. Ans. ~/625. 
4, Reduce a’ to a radical of the 5th degree. 


Ans. &/ a", 
5. Reduce — 5c to a radical of the 3d degree. 
| Ans. 3/—125e. 

6. Reduce — 3 to a radical of the 4th degree. 


Ans. 4/81 x. 


246 ELEMENTARY ALGEBRA. 


7. Reduce a+ x to a radical of the 3d degree. 
Ans. Sa? + 3a°x + 3a2?+ 2, 
8. Reduce 3¢-+ m to a radical of the 2d degree. 
Ans. /9 e+ 6em + m2. 
9. Reduce —2y to a radical of the 2d degree. 


Ans. J 4y’. 
10. Reduce — 2y toa radical of the 3d degree. 
Ans. /—8y. 


11. Express a./«x entirely under the radical sign. 


Ans. fax. 
12. Express a?./a—~w entirely under the radical sign. 


| Ans. Ja’ —a'e. 
13. Express 5,/8 entirely under the radical sign. 


3 Ans. 4/200. 

14. Express 6a’m%/a'm entirely under the radical sign. 
Ans. 3/216 a8m'. 

15. Express 2ecry 4/2erz entirely under the radical sign. 


Ans. ¥326r xy, 


Note. We know that 3 has no square root. No number multi- 
plied by itself produces 3. What we call an approximation to the 
square root of 3 is really the exact root of some number very near 3. 


Thus, 
ay nt Ao 


1.738 = 7/2.9929., 
1.782 = 7/ 2.999924. 
Such an expression as 7/3 is sometimes called an irrational quantity 
or a surd, because of its absurdity. Most numbers have no roots, 


of any degree. 


ELEMENTARY ALGEBRA. 247 


SECTION LXVI. 


Reduction of Radicals to their Simplest Forms. 


A RADICAL quantity is said to be reduced to its simplest 
form when as little of it as possible is left under the radical 
sign. We apply the principle: To take any root of a quan- * 
tity 1s the same as to take the same root of each of its factors 
and multiply the results together. 


Prosiem 1. Reduce /72a°%cx’ to its simplest form. 


Sotution. The greatest square factor of the quantity 
under the radical sign is 36a*zx”. The square root of this 
quantity may be taken, and the sign removed. Thus: 


J72a%cx? = 4/36 ate? x 2Qac= bax /2 a8. 
Prosiem 2. Reduce 7 4/16 a}. /ab? to its simplest 
form. 


SoLution. The quantity of which the 4th root is indi- 
cated is not a perfect 4th power, but is a perfect square. Its 


square root being taken, the radical becomes 7/4 ab". J ab? 
or 7V 4a°b®. This becomes 7/4 a°b® x b = 14.a°h*/0. 


Prosiem 3. Reduce 5ex Ycx(e—-z) to its simplest 
form. 

Sotution. Finding the greatest cube factor under the 
radical sign, we have 


Sex? /ex(e—x) x a(e—x) = 5 ea*(e— z) Y ?(e — 2x) 


EXAMPLES. 
Reduce the following radicals to their simplest forms. 
mV AVA Ans. 2 atin 
. /125. Ans. 5 4/5. 
. /124, Ans. 2 /31. 
ALE Ans. 3 /11. 
. /128. Ans. 8 /2. 


or em 0 be 


ELEMENTARY 


. Lldabe V 15076". 
. 8px S36 px. 
.5e Je 

. abe Jarry. 

525 / 28. 

yes be 

. om 41620 ac*m. 
. 18 axe /768 ay, 
. 0/1800. 

. L260 aty, 

. S20 + 4aba + 20%. 
Ope VOne 

. am Vf882cx. 

By V5 yz. 

8a fa?+Qab +B. 
. /108. 

METS 

. 0/482, 

sua/ 1D. 

. 64 akc, 

. ¥729mn. 


ALGEBRA. 


Ans. & f/2. 
Ans. m Jf/n. 
Ans. x? Jabe. 
Ans. 3m Jax. 
Ans. 2c /6ab. 
Ans. 3mn JV az. 
Ans. 5a? 2. 
Ans. 20x? /y. 
Ans. 15 76’c? ./15. 
Ans. 48 p’x /x. 
Ans. 25¢ fe. 


Ans. abe / xy. 
Ans, 125, 


Ans. a 
Ans. 18 em J/5da. 


Ans. 288 acx? /3 y. 
Ans. 150 ./2. 


Ans. 25a J/2 xy. 
Ans. (a+b) /2a. 
Ans. 6 pd /6. 
Ans. 2lam J/2ex. 
Ans. 1d y? /2. 
Ans. 3a7+ 3ab. 
Ans. 2° fx + y. 
Ans. 3.272 

Ans. 2 aie 


Ans. 6 8/2. 
Ans. 5 «7/3. 


Ans. 4a ec’, 
Ans. 9 Ymn. 


ELEMENTARY ALGEBRA. 249 


36.2 Y1024a2. Ans. 16 3/2z. 
87. Yeni — nt. Ans. n */e—n. 


88. a Y(a—x)ty. Ans. a(a—x) Y(a—a)y. 
39. m(n—1) Ye’. Ans. em?(n—1). 
40, 21 3/320 (a—z). Ans. 84 9/5 (a—z). 
41, 2/8000 mnt. Ans. 20n 3/min. 
42, 4/512 m'x*. Ans. 4a 4/2 m?. 


43. Y64a*m’, Ans. 2a /2am. 
44, </96 (a—x)*. Ans. 2(a—ax) ¥3(a—z). 


45.2 /8a(e—x)*. Ans. 4(e—2x) Ya’. 

46. 5 ab? 2/128 a’. Ans. 10 ab? 8/2 a. 

Ad. 62/8. Ans. 16. 

48. 16 2/16. Ans. 32 3/2. 

49. 64 4/64. Ans. 128 ./2. 
—cotg¢ oo —— 


SECTION LXVII. 


Elimination of Radicals from the Denominators 
of Fractions. 


2 
ProBiEM 1. Reduce 12 \/ 3 to an expression not hav- 


ing a radical denominator. 

SotuTion. Make the denominator a perfect square by 
multiplying it by itself or any other number that will 
answer. Of course, the numerator must be multiplied by 
the same number, that the value of the fraction may not 
be altered. 


2 [2x3 _ 6 


Now take the square root of the denominator and place 
it out of the radical sign. 


/6 1 
12 \/ea2x tve=4 J/6. 


250 ELEMENTARY ALGEBRA. 


ProsiEM 2. Reduce mes to an expression not hay- 
ing a radical denominator. | | 

Sorurion. Here squaring the denominator would give 
16—8 5+ 5, or 21—8 ./5, a radical expression. Hence 
some other plan must be adopted. The denominator is the 
difference of two numbers, and multiplying the difference 
of two numbers by their sum gives the difference of their 
squares. 


B+V5, 44+V5_124+7V54+5_ 174775 


A—/f/5 -44+/75 16-5 °° &£zx=£11 
3 
PROBLEM 3. Reduce Ui . to an expression not having 








a radical denominator. 
Sotutron. Make the denominator a perfect cube, and 
extract the cube root of it. 


aae SEEN fis 
Ja=V ran Va Ye 


EXAMPLES. 


Reduce the following expressions to others not having 
radical denominators. 


3 1 

I. Jee Ans. 5 16. 

2. /* Ans. Ye 
x 4 07 
1 1 

S Rh . Ans. gV¥2. 
3 | 

4, \/ r Ans. Vv oO 
9 3 

5. his. Ans. “iv 1: 


5 1 
6. \/ 7 Ans, 9/35. 


22. 


23. 


ELEMENTARY ALGEBRA. 251 





18 \/>- Ans. 2/15. 
3 5 
eo afm Ans. qv. 
vy) J Ans. 2/7. 
6 ~. Ans. aie 
. mn \ =, Ans. m /amn. 
. Vabe \/ a - Ans. 5 abe, 
2 
; f=. Ans. Sn. 
5a ss /35 ax 
‘ 72 ns. nda 
3/1 ; 1 ; 
P \ 2° Ans. a V4 
8 2 1 P 
5° Ans. 5 00. 
; 2 Ans. ave 
8 1325 1, 
we Ans. 9 v 680. 


tf 4 28 
at ee: Ans. y Set 
cy cy 


/*. ule: Mea 
nee Zs 
[p—q Ane VOR ODIED 
Be ae 3 
7— V3 PN pita enn 0 
5—/5 Pee 38— V9 


252 ELEMENTARY ALGEBRA. 








24. o4+ 73° Ans. 11 —6 Jd. 
5+ /2 | 

25. Bey 97 Ans. 19 -b 13 Jas 
1+ V3 

26. Raa Ans. 2 + Jd. 

Sih tee TA Ans li 
a— f/x a—ax 

i 3— fT 

er ey Hoi! Pewee 

28. Ji-3 ns 5 
5+4/2 30 + 28 f/2+15/5+4 12 410 

29. fas 75, Ans. ———— a —_—_———__—., 

ee Ju, 2 
m+ /n m—n 

—0$-¢.0-0-——_——_ 


SECTION LXVIII. 


Reduction of Radicals to a Common Index. 


ProspitEM. Reduce v/a, c’, and 4/5 to a common 
index. 
SotutTion. Expressing the radical quantities with frac- 


1 
tional indices, we have a, ce’, and BF. Reducing the frac- 
tional indices to others having the least common denomi- 
nator, we have xu, #3, and 7, instead of 4, 4, and +. The 

4A 
radical quantities thus become a’’, oft, and Bt, which may 
be expressed ¥/a', ¥/c®, and 4/125. » 
EXAMPLES. 
Reduce the following radicals to others having a common 
index. | 
Lisd/a, /a. Ans. </a?, a’. 
eed. 2/2. Ans. ¥/125, £/4. 


ELEMENTARY ALGEBRA. 253 


3. ft, “fa, </n. MARNE OVS ae Ri: 

Pee A/S 2, 6/7". Ana} '£/ a4) 8/20) 8/y", 

haa, a, —/ a. Ana orsar, Shae 
& 5 

6. m‘, o8, Ze, Ans <7, SS ON, 

7. Jate, Yam Y(r—1). 


Ans. S/(ate), Yaim, Y(r—1)’. 
8. </49, &/27, 2/512. Ans. “/7, /9, </16. 


Nort. In this example the indices can first be reduced by taking 
the square root of 49, the third root of 27, and the ninth root of 512. 
The radical quantities then become 7/7, 7/3, and 21. 


Oe tvat sr /(a—x)*. Ans. (a+ 2), f(a—ax)*. 
10. 7(e—n)s, Ve—x. Ans. Yo—n, Ye—x, 


SECTION LXIX. 


Addition and Subtraction of Radicals. 


Prostem. Find the value of 7 \/ +—5 of : 4. 


a3t2 /90. 


SoxtutTion. First reduce the radicals to their simplest 
forms with rational denominators. 


3 UN ay 6 Ta /i=! 
rs" me a AMER 

2 2xD 
aha ON BB a —5 \/3,=— v0. 


BBall) BiX2/Biyui BB 
= iV BK/S 1B 
22 





254 ELEMENTARY ALGEBRA. 


It must be remembered that none but similar quantities 
can be united into one term. 
$/3— +10 
/3+6 /10 
49 ./3 +5 ./10, the sum required. 





EXAMPLES. 


Unite the following expressions into as few terms as pos- 
sible: 


1. 162 + ./98 + /2. Ans. 17 2. 
2. /24— 150 + 54. Ans. 0. 
8. /15— f/27 — 3. ' Ans. /3. 
4,2./72—5 YV7—8 YV8 +468. Ans. 6/2477. 
5. 3 /4— /12584+ 4 YH. Ans. —2 /5. 
6. 38 f/2—2 v2. Ans. 3 /2—2 V2. 
7. 18./24+4 /27. Ans. 138 /2+4+ 12 /3. 
8. m Yem™*—a Yea’. Ans. (m?— a’) ¥/e. 
9. Y—729 + 7/2742 /48. Ans. —9+9 3. 
10. 28 3—5 4/108. Ans. — #/4. 
—-08g400——. 


SECTION LXX. 
Multiplication and Division of Radicals. - 


WE apply the principle: To take any root of a quantity is 
the same as to take the same root of each of its factors and mul- 
tuply the results together. 


ProBLEM 1. Multiply 5 4150 by 3 ./48. 

Sotution. Reduced to their simplest form, the radicals 
become 25 ./6 and 12 ./3. Multiplying these together, we 
have 300 ./18, which may be reduced to 900 4/2. 


ELEMENTARY ALGEBRA. 255 


PrRoBLEM 2. What is the value of </x es /a? 


Sotution. Reducing the radicals to a common index, 
2 


6 
° J J OG 
we have, 4/2’ and ¥/a*, Dividing, we have, y/ —. Re 
a 


3 
ducing this so as to have a rational denominator, we have, 
2x Ya £/ x? 

ix Se a 





EXAMPLES. 


Perform the operations indicated in the following ex- 
amples : 


Pate X/ 2. | Ans. X. 
Woes KX s/t. Ans. Jax. 
a. fd X 40. ~ Ans. 2 /30. 
eds ah 2 a/ 10. Ans. 30 /3. 
5..5 /15 + 5. Ans. 5 /3. 
6.12 ./32 + 4/288. Ans. 1. 
7. /40 + 8. Ans. /5. 
8.¢/axmVJ/n. fe Ans. cm Van. 
9, he x es Ans. 1. 
x c 


10. Vg oe yi sil ie 
x Cc 4 


Pree x </t X. 1/1. Ans. + 2/9. 
a2? xX </3. Ans. &/72. 
13. (fa + Sy) (/x— Vy). Ans. «— y. 
14. (fa + Sy) (a+ Vy) Ans. +2 J/xy + y. 


15. (/2— Sy) (2+ Say t y). Ans. © f/x—y SY. 





256 ELEMENTARY ALGEBRA, 


SECTION LAX 
An Algebraic Puzzle. 


Let the letter x stand for 2. Then we haye 


x og, 
Multiplying by 2, OO ee 
Subtracting 4, xv?—4 = 2x—4, 
Factoring, (x + 2) (x —2) = 2 (x — 2). 
Dividing by x — 2, e+ 2 = 2. 
Substituting 2 for x, 2+2=2. 
Or 4= 2, 


Now as this result is absurd, there must be some error in 
the operation. Yet it seems to have been conducted strictly 
according to algebraic principles. Hence the puzzle. 

The mystification may be easily cleared up. The false 
step is the division by x —2. Since & is equal to 2,2 —2 
is equal to 0; that is, it has no value. Now if we multiply 
nothing by any number whatever, the result is always 
nothing. 

We may have 6X0 == 5 x10; 


but we cannot divide by 0, and obtain 
G = 0: 
The equation 
(x + 2) (x —2) = 2 (x —2) means (x + 2) X O=2 x 0. 
This is true; but we cannot divide by 0, and obtain 
c+2 = 2. 


THE END. 

















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